If we were allowed to plug in $a=0$, we would find $f(1)=1$. With $n=1$, we then get $f(f(m))=m$, so $f$ is bijective. Unfortunately, we are not allowed to plug in $a=0$, but it does put us on the right track.

We start by plugging in $n=1$. This gives us $f(f(m) f(1))=m$, which means that $f$ is bijective. Indeed, if $f(m)=f(n)$, then we directly get $m=f(f(m) f(1))=f(f(n) f(1))=n$ and for every $m$ there is an $a$ such that $f(a)=m$, namely $a=f(m) f(1)$. Now we plug in $m=n$, which gives us $f(f(m)^2)=m^2=f(f(m^2) f(1))$. Due to the injectivity of $f$, it follows that $f(m)^2=f(m^2) f(1)$. Choose now $m_1$ such that $f(m_1)=1$. (From our proof of surjectivity, we can choose $m_1=f(1)^2$, but that is not important.) If we plug in $m=m_1$ in the above, we get $1=f(m_1^2) f(1)$. In particular, $f(1)$ is a divisor of 1, and since it must be a natural number, we conclude that $f(1)=1$.

If we now plug in $n=1$ again, we find that $f(f(m))=m$. Now we can plug in $f(m)$ and $f(n)$ for $m$ and $n$ so that we get the more classical equation

$$f(m n)=f(f(f(m)) f(f(n)))=f(m) f(n)$$

This means that $f$ is determined by the values $f(p)$ with $p$ prime.
Suppose we have an arbitrary prime number $p$, and let $m$ and $n$ be natural numbers such that $f(p)=m n$. It follows that $f(m) f(n)=f(m n)=f(f(p))=p$. Since $f(m)$ and $f(n)$ are natural numbers, it must be that $f(m)=1$ or $f(n)=1$. Due to injectivity, the only $m$ for which $f(m)=1$ is the number $m=1$. Therefore, $m=1$ or $n=1$. We conclude that $f(p)$ is a prime number.

Now suppose that $p \nmid 2024$. In particular, this means that $\operatorname{gcd}(p, 2024)=1$, so $p$ has a multiplicative inverse modulo 2024. Let $b$ be a natural number in the residue class of the inverse, i.e., $b p \equiv 1 \bmod 2024$. Since $b p \geq p > 1$, it follows from the second equation that $f(b p)=b p$, but also that $f(b p + 2024 p)=b p + 2024 p$. From the previous paragraph, however, it follows that $f(b p)=f(b) f(p)$ and $f((b+2024) p)=f(b+2024) f(p)$. This means that $f(p)$ is a divisor of both $b p$ and $b p + 2024 p$. Now we calculate that

$$\operatorname{gcd}(b p, b p + 2024 p) = \operatorname{gcd}(b p, 2024 p) = \operatorname{gcd}(b, 2024) p = p,$$

since $b$ is the multiplicative inverse of $p$ modulo 2024. Due to injectivity, we know that $f(p) \neq 1$, so $f(p) \mid p$ implies that $f(p)=p$.

Suppose for a $p \mid 2024$ that $f(p)=q$, with $q \neq p$ another prime number. Then it follows that $f(q)=f(f(p))=p$. So all prime numbers that do not map to themselves by $f$ form pairs $(p, q)$ with $f(p)=q$ and $f(q)=p$. Since $2024=2^3 \cdot 11 \cdot 23$, we have four solutions:

- $f(p)=p$ for all prime numbers $p$. So $f$ is the identity,
- $f(2)=11, f(11)=2$ and $f(p)=p$ for all $p \notin \{2,11\}$,
- $f(2)=23, f(23)=2$ and $f(p)=p$ for all $p \notin \{2,23\}$,
- $f(11)=23, f(23)=11$ and $f(p)=p$ for all $p \notin \{11,23\}$.

It is easy to check that these functions also satisfy the conditions.