Iranian Mathematical Olympiad

The answer is Yes! Let $d(n)$ be the number of divisors of natural number $n$. We want to construct $f$ such that for any positive integer $m$, $f(d(f(m)))=d(m)$. Let $A_k=\{n\in \mathbb{N}\mid d(n)=k\}$. For example, $A_1=\{1\}$ and $A_2$ is the set of prime numbers. Note that $A_k$ has an infinite number of elements for every $k>1$, because $p^{k-1}\in A_k$ for every prime number $p$. To define $f$, set $f(1)=1$, $f(2)=2$, $f(3)=5$ and $f(5)=3$. For each $n\ge 4$, suppose that $f(k)$ is defined for $1\le k\le n-1$. If $f(n)$ is not defined then let $j=f(d(n))$. $j$ is well defined because $d(n)<n$. Let $t$ be the least element of $A_j$ that $f$ has not been defined on it yet, so we have $d(t)=j$. Define $f(n)=t$ and $f(t)=n$. Therefore, for each natural number $n$, these properties are gained inductively: $$\begin{array}{rl}f(d(n))& =j=d(t)=d(f(n)),\\ f(f(n))& =n,f(f(t))=t,\\ f(d(t))& =f(j)=f(f(d(n)))=d(n)=d(f(t)).\end{array}$$ Hence for every $m\in \mathbb{N}$, we have $f(d(f(m)))=f(f(d(m)))=d(m)$.