Substituting $m = n$ gives $f(n) \mid f(n) - n$, so for all natural numbers $n$ we have $f(n) \mid n$. Applying this to the original condition, it follows that $f(n) \mid f(m)$ if and only if $n \mid m$.

We show that $f(n) = n$ by induction on the number of prime factors of $n$. The base of the induction is the case $n = 1$. In this case, we have $f(1) \mid 1$, therefore $f(1) = 1$.

Suppose that $f(k) = k$ for all natural numbers $k$ with fewer prime factors than $n$, and suppose for a contradiction that $f(n) \mid n$ is a strict divisor of $n$. Then there exists a prime number $p$ such that $f(n) \mid \frac{n}{p} = f(\frac{n}{p})$, by the induction hypothesis. But $n$ does not divide $\frac{n}{p}$, which contracts our assumption that $f(n) \neq n$. Therefore $f(n) = n$, and this concludes the induction.

Note that $f(n) = n$ is indeed a solution, since $n \mid m$ holds if and only if $n \mid m - n$. $\square$