Solution:
We first prove that $f$ is surjective. This follows directly from the second condition since for all $a \in \mathbb{N}$ we have
$$b=\underbrace{f(f(\cdots f(f}_{f(a)-1}(a)) \cdots)) \Longrightarrow f(b)=a$$
Now pick any prime $p>2021$ and $n$ with $f(n)=p$. Now apply $f$ to the second condition to get
$$p=f(n)=\underbrace{f(f(\cdots f(f}_{p+1}(n)) \cdots))=\underbrace{f(f(\cdots f(f}_{p}(p)) \cdots))$$
Instead setting $n=p$ in the second condition we find that
$$p=\underbrace{f(f(\cdots f(f}_{f(p)}(p)) \cdots))$$
as well. Now let $d_{p}$ be the smallest positive integer such that $p=\underbrace{f(f(\cdots f(f}_{d_{p}}(p)) \cdots))$.
It follows that $d_{p} \mid p, f(p)$ since otherwise we would obtain a smaller value of $d_{p}$ by repeatedly substituting this expression in the two equations above. Now if $d_{p}=1$ we have $f(p)=p$ and if $d_{p}=p$ we have $f(p)-p<2021<p$ using the first condition and we find $f(p)=p$ as well. We conclude that every prime $p>2021$ satisfies $f(p)=p$ and since there is an infinite number of such primes, this solves the problem.