Let $x_1,x_2,...,x_n$ are roots of $P(x)=x$ $x_i$ is also root of $P(P(P(x)))=P(x)$ so $x_i$ is real Let $x_i=x_j$ for some $i<j$ $(x-x_i)^2| P(x)-x$ then $(x-x_i)^2|P(P(x))-P(x) \to (x-x_i)^2 | P(P(P(x)))-P(P(x)) \to (x-x_i)^2 | P(P(P(x)))-P(x)$ But we have that all roots of $P(P(P(x)))-P(x)$ are distinct, so all roots of $P(x)=x$ are distinct real too. Now let $y_1,...,y_{n^2-n}$ are roots of $P(P(x))=x$ but are not roots of $P(x)=x$ Then with same logic we can prove that $y_i$ are real and distinct roots. Let $z_1,...,z_{n^3-n^2}$ are roots of $P(P(P(x)))=P(x)$ that are not roots of $P(P(x))=x$. They are all real and distinct. Let $P(x)=ax^n-asx^{n-1}+Q(x)$ where $deg (Q(x)) \leq n-2$ Then $x_1+x_2+...+x_n=s$ Not hard to show that $P(P(x))-x=a^nx^{n^2}-a^n snx^{n^2-1}+Q_2(x)$ where $deg(Q_2(x)) \leq n^2-2$ So $y_1+y_2+...+y_{n^2-n}=sn-s$ And also $P(P(P(x)))-P(x)=a^{n^2}x^{n^3}-a^{n^2}sn^2x^{n^3-1}+Q_3(x)$ where $deg (Q_3(x)) \leq n^3-2$ $z_1+z_2+...+z_{n^3-n^2}=sn^2-sn$ Then $\frac{x_1+x_2+...+x_n}{n}=\frac{s}{n}, \frac{y_1+...+y_{n^2-n}}{n^2-n}=\frac{s}{n}, \frac{z_1+z_2+...+z_{n^3-n^2}}{n^3-n^2}=\frac{s}{n}$

Let $Q(x) = P(P(P(x))) - P(x)$ , $R(x) = P(P(x)) - x$. Note that all roots of $R(x)$ are also roots of $Q(x)$. Let $r_1 , r_2, \dots , r_{n^2} \in \mathbb{R}$ be the roots of $R(x)$ and let $r_1 , r_2 , \dots , r_{n^3} \in \mathbb{R}$ be the roots of $Q(x)$. For $a_n \neq 0$, let $P(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ and let $t=\frac{-a_{n-1}}{a_n}$. By some computation, we get $R(x) = P(P(x)) - x = a_n^{n+1}x^{n^2} + a_n^nna_{n-1}x^{n^2 - 1} + A(x)$ where $deg (A(x)) \leq n^2-2$. So we get $r_1 + r_2 + \dots + r_{n^2} = tn$. Again after some computation we have $Q(x) = P(P(P(x))) - P(x) = a_n^{n^2 + n + 1}x^{n^3} + a_n^{n^2 + n}n^2a_{n-1}x^{n^3 - 1} + B(x)$ where $deg (B(x)) \leq n^3-2$. So we get $r_1 + r_2 + \dots + r_{n^3} = tn^2$. Finally, it is easy to see that $\frac{r_1 + r_2 + \dots + r_{n^2}}{n^2} = \frac{r_{n^2 + 1} + \dots + r_{n^3}}{n^3 - n^2} = \frac{t}{n}$ as desired.$\blacksquare$