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:
$$\begin{align*} f(d(n)) &= j = d(t) = d(f(n)), \\ f(f(n)) &= n, \quad f(f(t)) = t, \\ f(d(t)) &= f(j) = f(f(d(n))) = d(n) = d(f(t)). \end{align*}$$
Hence for every $m \in \mathbb{N}$, we have $f(d(f(m))) = f(f(d(m))) = d(m)$.