Firstly, we prove by induction that if $f$ is a regular function then $f(i) < i$ for each positive argument $i$. For that, assume that the claim holds for all smaller arguments. We have $f(i) \in \{i-1, f(i-1), f(f(i-1)), \dots\}$ by regularity. By the induction hypothesis and the assumption $f(0) = 0$, we have $i-1 \ge f(i-1) \ge f(f(i-1)) \ge \dots$. Hence $f(i) < i$ indeed.

Now we prove by induction on $n$ that the information given to Miku by Juku uniquely determines the regular function. If $n = 0$ then only one regular function exists whence the claim holds trivially. Assume in the rest that $n > 0$ and that the claim holds for all smaller numbers. As $f(1) < 1$ by the lemma proven in the beginning, we must have $f(1) = 0$.

Let $k$ be the largest number in $\{0, 1, \dots, n\}$ such that $f(k) = 0$. Then $k \ge 1$ since $f(1) = 0$. By the lemma proven in the beginning, $0 < i \le k$ always implies $f(i) < i$ and hence also $f(i) < k$. On the other hand, $k < i \le n$ implies $f(i) \ge k$. Indeed, assume that $f(j) \ge k$ for each $j = k+1, \dots, i-1$; let exactly $s$ initial members of the sequence $i-1, f(i-1), f(f(i-1)), \dots$ be larger than or equal to $k$. Then the $s$th member must be $k$ because otherwise the $s$ initial members would all be in $\{k+1, \dots, i-1\}$, implying that the next member would be larger than or equal to $k$, too. Thus the next member is $f(k)$ which equals 0 and all following members are 0, too. But as $f(i) \ne 0$, we must have $f(i) \ge k$. We can conclude that the number of arguments $i$ such that $f(i) \ge k$ is exactly $n-k$.

Now if $0 < l < k$ then similarly to what we did before we can prove that $0 < i \le l$ always implies $f(i) < l$. But among the arguments $i$ such that $l < i \le n$, there is $k$ for which $f(k) = 0$. Thus the number of arguments $i$ such that $f(i) \ge l$ is less than $n-l$. Consequently, $k$ is the least positive integer for which the number of arguments $i$ such that $f(i) \ge k$ is $n-k$. According to this property, Miku can determine $k$.

Restricting $f$ to $\{0, 1, \dots, k-1\}$, all conditions of regularity hold. Hence, by the induction hypothesis, Miku can determine $f(1), \dots, f(k-1)$. Note that if we forget about arguments $1, \dots, k$ and decrease the other argument-value pairs of $f$ by exactly $k$, the conditions of regularity again hold. Hence, by the induction hypothesis, Miku can also determine $f(k+1), \dots, f(n)$. Consequently, Miku can determine the values of $f$ at all its arguments.

We start like in Solution 1 by showing that if $f$ is regular then $f(i) < i$ for every positive argument $i$.

Now we prove the following lemma: Whenever $0 < k \le i \le n$, either $f(i) \ge k$ or $f(i) \le f(k)$. To this end, fix $k$ and proceed by induction on $i$. The base case $i = k$ is trivial. Assume now that $i > k$ and, for every $i'$ such that $k \le i' < i$, either $f(i') \ge k$ or $f(i') \le f(k)$. By regularity, $f(i)$ is a term in the sequence $i-1, f(i-1), f(f(i-1)), \dots$. This sequence decreases until it reaches zero. Let the sequence contain exactly $s$ terms larger than or equal to $k$. As all terms are less than $i$, the induction hypothesis implies that the term number $s+1$ is at least $k$ (which is impossible by the choice of $s$) or at most $f(k)$. Then the following terms are also at most $f(k)$. Thus $f(i)$ must be either at least $k$ or at most $f(k)$.

Miku can determine Juku's regular function as follows. Let Miku put the numbers of repetitions of values in the order of decreasing value (i.e., the number of repetitions of $n$ first, then the number of repetitions of $n-1$, etc.). Every time a number of repetitions of some value $k$ equals a positive number $a$, let Miku define the value of $f$ on the least $a$ arguments greater than $k$ where the value is not defined so far to be $k$. We show that this results in the only possible regular function with the given numbers of repetitions of values. Suppose the contrary. Let $k$ be the largest value of the function for which a divergence appears. We know that all arguments on which the function obtains the value $k$ must be larger than $k$. Hence there exist $i'$ and $i$ such that $k < i' < i$, $f(i) = k$ and $f(i') \neq k$. Since all values greater than $k$ are already assigned by the algorithm and they are correct by the choice of $k$, the only option is $f(i') < k$. But then the lemma proven above implies either $f(i) \ge i' > k$ or $f(i) \le f(i') < k$, contradiction. Consequently, no divergence between the actual function and the one constructed by the algorithm can arise. Hence Miku can always determine Juku's function.

Like in Solution 1, we show that if $f$ is regular then $f(i) < i$ for every positive argument $i$. Like in Solution 2, we show that $0 < k \le i \le n$ always implies either $f(i) \ge k$ or $f(i) \le f(k)$.

Miku can determine Juku's regular function as follows. Let Miku start by defining $f(0) = 0, f(1) = 0$. These are clearly the only options. Whenever $f(0), f(1), \dots, f(i-1)$ have been fixed, let Miku define $f(i)$ to be the largest number among $0, 1, \dots, i-1$ whose presumed number of occurrences is not achieved yet. As $f(i) < i$, such number must exist. We show now that this definition is the only possible. Suppose the contrary. Let the first divergence arise at the definition of $f(i)$. Let the algorithm assign the value $k$ to it; then actually $f(i) < k < i$, because the algorithm finds the largest number less than $i$ that can be used. But since the value $k$ is not yet saturated, there must exist $i'$ such that $i' > i$ and $f(i') = k$. By the lemma proven at the beginning of the solution, either $f(i') \ge i > k$ or $f(i') \le f(i) < k$, contradiction. Hence Miku can always determine Juku's function.

Now we prove by induction on $n$ that the information given to Miku by Juku uniquely determines the regular function. For $n=0$, there exists only one regular function, whence the claim holds. Assume in the following that $n > 0$ and the claim holds for $n - 1$. Let $a_0, a_1, \dots, a_n$ be the numbers of repetitions of values of $f$. By the property proven at the beginning of the solution, $n$ cannot occur among the values of $f$, hence $a_n = 0$. Let $l$ be the least positive integer such that $a_l = 0$. As $a_{l-1} > 0$, we have $f(l) = l - 1$ by the above. Define $b_0, b_1, \dots, b_{n-1}$ by
$$b_i = \begin{cases} a_i, & \text{if } i < l - 1, \\ a_i - 1, & \text{if } i = l - 1, \\ a_{i+1}, & \text{if } i \ge l. \end{cases}$$
Note that $b_0, b_1, \dots, b_{n-1}$ are the numbers of repetitions of values of function $g$ defined by
$$g(i) = \begin{cases} f(i), & \text{if } i < l, \\ f(i+1), & \text{if } f(i+1) < l \le i, \\ f(i+1) - 1, & \text{if } l < f(i+1). \end{cases}$$
In other words, function $g$ is obtained from $f$ by removing the argument-value pair $(l, l-1)$ (this removes the only occurrence of $l$ in these pairs) and decreasing the numbers $l+1, \dots, n$ by 1 in all other pairs. It is easy to check that regularity of $f$ implies regularity of $g$. By the induction hypothesis, $g$ is the only regular function with numbers of repetitions of values are $b_0, b_1, \dots, b_{n-1}$. But $f$ can be defined in terms of $g$:
$$f(i) = \begin{cases} g(i), & \text{if } i < l, \\ l-1, & \text{if } i = l, \\ g(i-1), & \text{if } g(i-1) < l < i, \\ g(i-1)+1, & \text{if } l \le g(i-1). \end{cases}$$
In other words, add the argument-value pair $(l, l-1)$ and increase all numbers $l, \dots, n-1$ by 1 in other pairs. This completes the solution.

Like in Solution 1, we show that if $f$ is regular then $f(i) < i$ for every positive argument $i$. Next we show that $f(i) = i-1$ whenever $f$ is regular and $i-1$ occurs among the values of $f$. Indeed, let $k$ be the least positive integer such that $f(k) = i-1$; by the above, $k \ge i$. By regularity, $i-1$ occurs in the sequence $k-1, f(k-1), f(f(k-1)), \dots$. If $i-1 = k-1$ then $f(i) = i-1$, otherwise $i-1$ can be represented in the form $f(j)$ where $j \le k-1 < k$, which contradicts the choice of $k$.

Now we prove by induction on $n$ that the information given to Miku by Juku uniquely determines the regular function. For $n=0$, there exists only one regular function, whence the claim holds. Assume in the following that $n > 0$ and the claim holds for $n - 1$. Let $a_0, a_1, \dots, a_n$ be the numbers of repetitions of values of $f$. By the property proven at the beginning of the solution, $n$ cannot occur among the values of $f$, hence $a_n = 0$. Let $l$ be the least positive integer such that $a_l = 0$. As $a_{l-1} > 0$, we have $f(l) = l - 1$ by the above. Define $b_0, b_1, \dots, b_{n-1}$ by
$$b_i = \begin{cases} a_i, & \text{if } i < l - 1, \\ a_i - 1, & \text{if } i = l - 1, \\ a_{i+1}, & \text{if } i \ge l. \end{cases}$$
Note that $b_0, b_1, \dots, b_{n-1}$ are the numbers of repetitions of values of function $g$ defined by
$$g(i) = \begin{cases} f(i), & \text{if } i < l, \\ f(i+1), & \text{if } f(i+1) < l \le i, \\ f(i+1) - 1, & \text{if } l < f(i+1). \end{cases}$$
In other words, function $g$ is obtained from $f$ by removing the argument-value pair $(l, l-1)$ (this removes the only occurrence of $l$ in these pairs) and decreasing the numbers $l+1, \dots, n$ by 1 in all other pairs. It is easy to check that regularity of $f$ implies regularity of $g$. By the induction hypothesis, $g$ is the only regular function with numbers of repetitions of values are $b_0, b_1, \dots, b_{n-1}$. But $f$ can be defined in terms of $g$:
$$f(i) = \begin{cases} g(i), & \text{if } i < l, \\ l-1, & \text{if } i = l, \\ g(i-1), & \text{if } g(i-1) < l < i, \\ g(i-1)+1, & \text{if } l \le g(i-1). \end{cases}$$
In other words, add the argument-value pair $(l, l-1)$ and increase all numbers $l, \dots, n-1$ by 1 in other pairs. This completes the solution.