Let $P(m,n)$ denote the assertion $f(f(m)f(n))=mn$. $P(m,1): f(f(m)f(1))=m$ so $f$ is injective. $P(mn,1): f(f(mn)f(1))=mn\implies f(f(m)f(n))=f(f(mn)f(1))\implies f(m)f(n)=f(mn)f(1)$ $m,n\mapsto 2023,4047: f(2023)f(4047)=f(2023\cdot 4047)f(1)\implies f(1)\implies f(1)=1$ $P(n,1): f(f(n))=n$. Thus, we have $f(mn)=f(m)f(n), f(f(n))=n$ for all $m,n$. By easy induction, we have $f(n^k)=f(n)^k$ so $f(n^{\varphi(2022)})=f(n)^{\varphi(2022)}\implies n^{\varphi(2022)}=f(n)^{\varphi(2022)}\implies f(n)=n$ for all $n$ relatively prime to $2022$ by Euler's. Will Continue Later

parmenides51 wrote: Let $N$ be the set of positive integers. Find all functions $f : N \to N$ such that both $\bullet$ $f(f(m)f(n)) = mn$ $\bullet$ $f(2022a + 1) = 2022a + 1$ hold for all positive integers $m, n$ and $a$. Let $c=f(1)$ $f(c^2)=f(f(1)f(1))=1$ $c^22023=f(f(c^2)f(2023))=f(f(2023)=f(2023)=2023$ and so $c=1$ So $f(f(n))=f(f(1)f(n))=n$ (and $f(n)$ is bijective) And $f(mn)=f(f(f(m))f(f(n)))=f(m)f(n)$ and $f(x)$ is multiplicative and totally defined by knowledge of $f(p)$ for all prime $p$ Let prime $p\notin\{2,3,337\}$ $p^{\varphi(2022)}\equiv 1\pmod{2022}$ $\implies$ $f(p)^{\varphi(2022)}=f(p^{\varphi(2022)})=p^{\varphi(2022)}$ and so $f(p)=p$ $f(f(2))=2$ and so $f(2)$ contains only primes $2,3,337$ $f(f(3))=3$ and so $f(3)$ contains only primes $2,3,337$ $f(f(337))=337$ and so $f(337)$ contains only primes $2,3,337$ And it is easy to conclude that $(f(2),f(3),f(337))\in\{(2,3,337),(2,337,3),(337,3,2),(3,2,337)\}$ And so $\boxed{f(n)=\frac n{2^{v_2(n)}3^{v_3(n)}337^{v_{337}(n)}}u^{v_2(n)}v^{v_3(n)}w^{v_{337}(n)}\quad\forall n\in\mathbb Z_{>0}}$, which indeed fits, whatever is $(u,v,w)\in\{(2,3,337),(2,337,3),(337,3,2),(3,2,337)\}$