I claim $f(n)=n$ for all $n$. Note that taking $m=n$, we find $f(n)\mid n$ for all $n$ and thus $f(n)\mid f(m)\iff n\mid m$. Now if $p$ is a prime, then $f(p)\in \{1,p\}$. Assume there is a prime $p$ such that $f(p)=1$. Then taking $m\equiv 1\pmod{p}$ and $n=p$, we get $f(n)\mid f(m)-n$ whereas $n\nmid m$. So, $f(p)=p$ for every prime $p$. We now show for every $n=p^k$, where $p$ is a prime and $k\ge 1$ is arbitrary, we have $f(n)=p^k$. We induct on $k$. The case $k=1$ is verified already, assume $f(p^{k-1})=p^{k-1}$. Taking $n=p^k$, we get $p^{k-1}\mid f(n)\mid n$, so $f(n)\in\{p^{k-1},p^k\}$. Now if $f(n)=p^{k-1}$ then taking $n=p^k$ and $m=p^{k-1}$ we find a contradiction. So $f(p^k)=p^k$. We are now ready for conclusion. Letting $N=\textstyle \prod_{i\le L}p_i^{e_i}$, where $p_1<\cdots<p_L$ are distinct primes, we find that $f(p_i^{e_i})\mid f(N)$ for all $i$. That is, $N\mid f(N)$. But as $f(N)\mid N$, we find $f(N)=N$, as requested.

Let $P(n,m)$ denote the assertion. $P(n,n): f(n)|f(n)-n \implies f(n)|n$ Now since $f(n)|n$, $f(n)|f(m)-n \iff f(n)|f(m)$. Hence $n|m \iff f(n)|f(m)$ If $f(a)=f(b)$, $f(a)|f(b)$ and $f(b)|f(a)$ so $a|b$ and $b|a$ $\implies a=b$, $f$ injective Now clearly $f(1)|1$, so $f(1)=1$. Now we will inductively show $f(n)=n$. If $f(i)=i$ for all $i<n$, then $f(n)\leq n$. If $f(n)<n$, $f(n)=f(f(n))$, so $n=f(n)$ contradiction. Hence $f(n)=n$ and we are done.