IRL ABooklet

Solution 1. We first prove inductively that the following equations hold for all $i \in \mathbb{N}$:
$$\begin{aligned} f(1 \cdot 4^{i-1}) &= 2 \cdot 4^{i-1}, \\ f(2 \cdot 4^{i-1}) &= 3 \cdot 4^{i-1}, \\ f(3 \cdot 4^{i-1}) &= 4 \cdot 4^{i-1} = 4^i. \end{aligned}$$
The base case follows easily from the string of inequalities
$$1 < f(1) < f(f(1)) < f(f(f(1))) = 4.$$
The inductive step then follows from the assumed identity for $f(f(f(n)))$.

Let $A = 1 \cdot 4^{i-1}$ for some $i \in \mathbb{N}$, let $S$ be the set of integers between $A$ and $2A$ inclusive, and let $T$ be the set of integers between $2A$ and $3A$ inclusive. Since $f$ maps the endpoints of $S$ to the endpoints of $T$, the strictly increasing condition for $f$ implies that
* $f(S) \subset T$, and
* $S$ and $f(S)$ have the same cardinality.
But $S$ and $T$ clearly have the same cardinality, so we must have $f(S) = T$, and monotonicity now implies that $f(A + k) = 2A + k$ for all $1 \le k \le 4^{i-1}$. In the same way, we see that $f(2A+k) = 3A+k$ for all $1 \le k \le 4^{i-1}$. Finally, $f(3A+k) = f(f(f(A+k))) = 4A+4k$ for all $1 \le k \le 4^{i-1}$. We now have a formula for $f(n)$ for all $n$.
Writing $2022 = 1024 + 998 = 4^5 + 998$, we see that
$$f(2022) = 2 \cdot 1024 + 998 = 3046.$$

Solution 2. Applying $f$ to $f(f(f(n))) = 4n$ and then using the same equation with $n$ replaced by $f(n)$ we get
$$f(4n) = f(f(f(f(n)))) = 4f(n)$$
for all $n \in \mathbb{N}$, and hence
$$f(4^n n) = 4f(4^{n-1} n) = \dots = 4^n f(n).$$
We know $f(1) = 2$, $f(2) = 3$, $f(3) = 4$ since the function is increasing with
$$1 < f(1) < f(f(1)) < f(f(f(1))) = 4.$$
Hence $f(4^i) = 4^i f(1) = 2 \cdot 4^i$ and $f(2 \cdot 4^i) = 4^i f(2) = 3 \cdot 4^i$ so
$$4^i \le \sum_{x=4^i}^{2 \cdot 4^i - 1} [f(x+1) - f(x)] = f(2 \cdot 4^i) - f(4^i) = 4^i.$$
Thus $f(x+1) - f(x) = 1$ for each $x$ with $4^i \le x \le 2 \cdot 4^i - 1$. It follows that $f(4^i + k) = f(4^i) + k = 2 \cdot 4^i + k$ for $1 \le k \le 4^i$ and hence $f(2022) = 2 \cdot 1024 + 998 = 3046$ as above.