Define $f(x)$ with $f:\Bbb R\to\Bbb R$. Does there exist a natural number $n>2$ such that $f^{(n)}(x)=f^n(x)$ never holds for any non-constant function $f(x)$ ?
Here $f^{(n)}(x)=f(f(f(\dots f(x))))$ and $f^n(x)=f(x)\cdot f(x)\cdot...\cdot f(x)$
Clearly $n=1$ is trivial. $n=2$ it becomes $$f(f(x))=f^2(x)$$ I found $f(x)=x^2$ is a solution.
$$f(f(x)=f(x^2)=x^4=((f(x))^2$$
I could not proceed for $n\ge 3$.
For any integer $n$ there is a nonconstant function $f$ such that $f^{(n)}(x) = (f(x))^n$ is true. Indeed, for $n \ge 2$ let $a_n$ be a positive real number such that $(a_n)^n = na$. Next, fix an $n$ and let $f(x) = |x|^{a_n}$. For example, $a_n = \sqrt[n-1]{n}$ will work e.g., $a_2=2$, $a_3 = \sqrt{3}$, $a_4 = 4^{\frac{1}{3}}$, $a_5 =5^{\frac{1}{5}}$, and so on and so forth.
Then on the one hand [for that particular $n$]:
$$f^{(n)}(x) = |x|^{(a_n)^n},$$
and on the other hand:
$$(f(x))^n = |x|^{na_n},$$
but with $(a_n)^n = na_n$ however, these two functions are the same.
You should add some continuity assumption. Otherwise here is a counterexample.
Let $f(x)=\chi_{(0,1]}(x)$, so $f^2(x)=f(x)^2=f(x)$. By induction, this implies that $f^n(x) = f(x)^n = f(x)$ for every $n$.
Without continuity assumption you can create infinite many counterexamples using the so-called simple functions, $f=\sum_{j=1}^mc_j\cdot\chi_{E_j}$ where $(E_j)_j$ are disjointed subsets. So $f^n(x)$ and $f(x)^n$ depend only on $c_1,\dots,c_m$ and $E_1,\dots,E_m$. You can make $c_j$ to be the roots of certain polynomials and $E_j$ to be arbitrary sets which contain or not contain certain $c_k$.
For example, take $E$ such that $-1,1\in E$ and take $f=\chi_E-\chi_{E^c}$, so $f^2(x)=f(x)^2\equiv1$. This is a case for $n=2$. If you take $1\in E$ and $-1\in E$ you get $f^3(x)=f(x)^3$ and then by induction $f^{2k+1}(x)=f(x)^{2k+1}$.
