From $f(f(n)) = n$ we obtain for all integers $k, n > 0$ that $n = f(k)$ if and only if $f(n) = k$.

We first show that $f(2n+1) = 2f(n)$ for all $n > 0$. To see this, let $k = f(n)$. Then $n = f(k)$ and $f(2k) = 2f(k)+1 = 2n+1$ and so $2f(n) = 2k = f(2n+1)$.

Next we show that $f(2^k n) = 2^k f(n) + 2^k - 1$ for all $k \ge 0$ and $n \ge 1$. We prove this by induction on $k$. When $k = 0$ this is obvious. For the inductive step we use the given $f(2n) = 2f(n) + 1$ to obtain
$$\begin{aligned} f(2^{k+1}n) &= f(2 \cdot 2^k n) = 2f(2^k n) + 1 \\ &= 2(2^k f(n) + 2^k - 1) + 1 = 2^{k+1}f(n) + 2^{k+1} - 1. \end{aligned}$$
Because $f(1) = 1$, we obtain $f(2^k) = 2^{k+1} - 1$. Using $f(f(n)) = n$ this gives $f(2^{k+1} - 1) = 2^k$ for all $k \ge 0$.

We can now work backwards to find $f(2020)$:
$$\begin{align*} f(2020) &= f(4 \cdot 505) = 4f(505) + 3 \\ f(505) &= 2f(252) \\ f(252) &= f(4 \cdot 63) = 4f(63) + 3 \\ f(63) &= f(2^6 - 1) = 2^5 = 32. \end{align*}$$
We finally obtain $f(252) = 4 \cdot 32 + 3 = 131$, $f(505) = 262$ and $f(2020) = 4 \cdot 262 + 3 = 1051$.

Look for a sequence $(a_1, b_1), (a_2, b_2), (a_3, b_3), \dots, (a_k, b_k)$ such that $(a_1, b_1) = (1, 1)$, $a_k = 2020$ and for $j \ge 2$, either $(a_j, b_j) = (b_{j-1}, a_{j-1})$ or $(a_j, b_j) = (2a_{j-1}, 2b_{j-1} + 1)$. The three given conditions for $f$ imply that for any such sequence $b_j = f(a_j)$ for $1 \le j \le k$. In particular, $f(2020) = b_k$. The following sequence works (ignore the fourth column for now)

$$f(n) = 3 \cdot 2^k - n - 1.$$

After showing that $f(2n+1) = 2f(n)$ as in Solution 1, it is possible to work out the first few $f(n)$ by hand. These values suggest that when $k \ge 0$ is an integer and $2^k \le n < 2^{k+1}$ we have
$$f(n) = 3 \cdot 2^k - n - 1.$$
We show this by induction on $k \ge 0$. When $k = 0$ we must have $n = 1$ and the claim is true, because $f(1) = 1$. For the inductive step we assume the formula is true for numbers strictly between $2^{k-1} - 1$ and $2^k$, for a given $k \ge 1$.

We consider two cases: firstly when $n$ is even, and secondly when $n$ is odd. If $n$ is even and $2^k \le n < 2^{k+1}$, then $2^{k-1} \le \frac{n}{2} < 2^k$ and so, by assumption,
$$f\left(\frac{n}{2}\right) = 3 \cdot 2^{k-1} - \frac{n}{2} - 1.$$
Substituting the doubling rule from the original question it follows that:
$$f(n) = 2f\left(\frac{n}{2}\right) + 1 = 3 \cdot 2^k - n - 1.$$
Thus, the result holds for even $n$. Now if $n$ is odd, $2^k \le n < 2^{k+1}$ implies
$$3 \cdot 2^k - 2^{k+1} - 1 < 3 \cdot 2^k - n - 1 \le 3 \cdot 2^k - 2^k - 1.$$
Simplifying, and using the fact that all three quantities are integers, this is equivalent to:
$$2^k \le 3 \cdot 2^k - n - 1 < 2^{k+1}.$$
Now as the middle term is even, we have earlier proved the formula for $f$ for even numbers in this range, which gives:
$$f(3 \cdot 2^k - n - 1) = 3 \cdot 2^k - (3 \cdot 2^k - n - 1) - 1 = n.$$
Then the self-inverse property of $f$ implies $f(n) = 3 \cdot 2^k - n - 1$. The induction step is now complete. Having proved this result, we finally note that:
$$2^{10} = 1024 \le 2020 < 2^{11}$$
and so we can immediately calculate $f(2020) = 1051$.