Observe that if $deg(p)\geq 2$ then by the mean value theorem $p(x+p(x))-p(p(x))$ grows like $x\cdot p'(c_{x})$ for some $c_{x}$ between $p(x)$ and $x+p(x)$. This cannot hold, because $p(x)$ grows much smaller. Thus the degree is 1 at most and then it's easy.

harazi wrote: Observe that if $deg(p)\geq 2$ then by the mean value theorem $p(x+p(x))-p(p(x))$ grows like $x\cdot p'(c_{x})$ for some $c_{x}$ between $p(x)$ and $x+p(x)$. This cannot hold, because $p(x)$ grows much smaller. Thus the degree is 1 at most and then it's easy. didn't understand, can anyone give a detailed answer?

An elementary solution: $ p(x)=\sum_{i=0}^{n}a_{i}x^{i}$,$ p(x+p(x))-p(p(x))=p(x)$, if $ n\ge 2$ on the left side we will have an element $ na_{n}^{n}x^{n(n-1)+1}$, while the right side has $ grad=n$, so $ n\le 1$, we get $ p(x)=ax$, $ a\in \mathbb{R}$

It is enough to solve this problem by noting that Q(x)=P(x)-x has only one root 0 or Q(x)=0