China Southeastern Mathematical Olympiad

If $n$ is odd, then all factors of $n$ are odd. So $f(n)$ is odd and $g(n)$ is even. $g(n)$ cannot be $f(n)$ to some power of positive integer. Therefore $n$ is even. The smallest two divisors of $n$ are $1$ and $2$, and the largest two divisors of $n$ are $n$ and $n/2$.

Let $d$ be the third smallest divisor of $n$. If there exists $k \in \mathbb{N}^*$ such that $g(n) = f^k(n)$, then

$$\frac{3n}{2} = (1 + 2 + d)^k = (3 + d)^k \equiv d^k \pmod{3}.$$

Since $3 \nmid \frac{3n}{2}$, we see that $3 \mid d^k$. So $3 \mid d$, and since $d$ is the third smallest, we see that $d = 3$.

Thus, $\frac{3}{2}n = 6^k$, we get $n = 4 \times 6^{k-1}$. Since $3 \nmid n$, we see that $k \ge 2$.

Summing up, $n = 4 \times 6^l \ (l \in \mathbb{N}^*)$.