, 2017

Let $d(n) = f(f(n))$ denote the number of divisors of $n$ and observe that $f(d(n)) = f(f(f(n))) = d(f(n))$ for all $n$. Also note that because all divisors of $n$ are distinct positive integers between $1$ and $n$, including $1$ and $n$, and excluding $n-1$ if $n > 2$, it follows that $2 \le d(n) < n$ for all $n > 2$. Furthermore $d(1) = 1$ and $d(2) = 2$.

We first will show that $f(2) = 2$. Let $m = f(2)$ and note that $2 = d(2) = f(f(2)) = f(m)$. If $m \ge 2$, then let $m_0$ be the smallest positive integer satisfying that $m_0 \ge 2$ and $f(m_0) = 2$. It follows that $f(d(m_0)) = d(f(m_0)) = d(2) = 2$. By the minimality of $m_0$, it follows that $d(m_0) \ge m_0$, which implies that $m_0 = 2$. Therefore if $m \ge 2$, it follows that $f(2) = 2$. It suffices to examine the case in which $f(2) = m = 1$. If $m = 1$, then $f(1) = f(f(2)) = 2$ and furthermore, each prime $p$ satisfies that $d(f(p)) = f(d(p)) = f(2) = 1$ which implies that $f(p) = 1$. Therefore $d(f(p^2)) = f(d(p^2)) = f(3) = 1$ which implies that $f(p^2) = 1$ for any prime $p$. This implies that $3 = d(p^2) = f(f(p^2)) = f(1) = 2$, which is a contradiction. Therefore $m \ne 1$ and $f(2) = 2$.

It now follows that if $p$ is prime then $2 = f(2) = f(d(p)) = d(f(p))$ which implies that $f(p)$ is prime. $\square$