My problem! Here are two solutions. Solution 1. Note that there are finitely many functions $g:S\to S$. Given $n$, let $g_n$ be a function so that $g_n^n=f$, and fix some positive integer $r>1$. Consider $$h_k=g_{r^k}$$for all $k\in\mathbb{Z}_{\geq 0}$. As $h_k$ can take on only finitely many values, there must be some $i\geq 0, j>0$ so that $h_i=h_{i+j}$. Call this function $h$. Then $$h^{r^i}=f=h^{r^{i+j}}\implies f=f^{r^j}.$$Let $s$ be minimal so $f^{1+s}=f$, and assume for the sake of contradiction that $s>1$. Taking $r=s$ gives $$f^{s^j}=f$$for some $j>0$. However, $$f^{s^j}=f^{s^j-1-s}f^{s+1}=f^{s^j-s}=f^{s^j-2s}=\cdots=f^s,$$so $f=f^s$, contradicting the minimality of $s$ and finishing the proof. Solution 2. Let $m=|S|$ and let $g$ exist so that $g^{m!}=f$. We claim that $g^{2m!}=g^{m!}$ for any function $g:S\to S$. Indeed, given some $x_0$, we have that $g$ takes $$x_0\to x_1\to\cdots\to x_{a-1}\to y_1\to y_2\to\cdots \to y_b\to y_1\to \cdots,$$ where the $\{x_i\}$ portion represents the portion before entering a cycle and the $\{y_i\}$ portion represents a cycle (if $x_0$ is in a cycle, then $y_1=x_0$). In particular, the size of the $\{x_i\}$ string must be $\leq m$ and the size of the $\{y_i\}$ portion must also be $\leq m$, so $a,b\leq m$. Thus, after applying $g$ on $x_0$ $m!\geq m$ times, the result is in a cycle, and, as $b\leq m\implies b|m!$, after applying it $m!$ more times, the result is in the same place in the cycle as it started. So, $g^{2m!}(x)=g^{m!}(x)$ for all $x$. This gives $f^2=f$.

Here’s the solution used in the proposal: First, since $S$ is finite, there is a finite number of functions $\{g_1, g_2, \dots, g_k\}$ from $S$ to itself. The function $f$ assigns to each positive integer $n$ one of these functions, and so it induces a partition of the positive integers into sets $P_i$ of all the integers assigned to $g_i$. By Schur’s theorem there are integers $a$, $b$ and $a+b$ in a single set $P_j$. Hence, some function $g\colon S \rightarrow S$ satisfies $f = g^a = g^b = g^{a+b}$, and so, $f^2= g^ag^b= g^{a+b}= f$.

Note sure if my solution is correct, please help check, thanks! Lemma: when S is finite, for a function g: S->S, there is a N and T to guarantee that when j>N, g^[j+T](x)=g^[j](x). This can be proved as below: Start from an element x0, list x0, g(x0), gg(x0), ..., since S is finite, eventually at some point we will see g^[i0+t0](x0)=g^[i0](x0). stop and restart the operation from another start element. Repeating this for all xj in S, we get ij and tj for each of the element, |S| pairs in total. Let N=max(i1, i2, ...i|S|), T=lcm(t1, t2, ...t|S|), then we have that for any x, j>N, g^[j+T](x)=g^[j](x). Back to the problem, since S is finite, the g: S->S is also finite (|S|^|S| possible mappings), get all the N for each g, pick the max one, denote it as N*; get all the T for each g, find a common multiple of them(not necessary to be lcm) to satisfy that T*>N*, denote it as T*, then via applying the lemma we have that for each g, when j>N*, g^[j+T*](x)=g^[j](x). Now, since f(x) is a T*-th power function, there is a g that f(x)=g^[T*](x), and we have T*>N*, then ff(x)=g^[T*+T*](x)=g^[T*](x)=f(x), which means the answer is yes, ff(x)=f(x) is true for each x.