Let $R(x) = \frac{F(x)}{G(x)}$ be a rational function with $F(x), G(x) \in \mathbb{Z}[x]$ and
$F(x), G(x)$ have no common root modulo $p$ for all primes $p$.
Consider the rational function $Q(x)=\underbrace{R(R(\ldots(R(x))))}_{\text{$n$ times}}$ where $n \in \mathbb{N}$.
Prove that if there is an integer $k$ such that $Q(k)=k$, then $R(R(k))=k$.
My thought,
in case $G(x) = 1$, $R(x)$ will be polynomial.
Let $a_1$ be the value such that $Q(a_1)=a_1$ and $R(a_i)=a_{i+1}, \forall i= 1, 2, \ldots, n-1$.
We have $a_2-a_1 \mid a_3-a_2 \mid \ldots \mid a_1-a_n \mid a_2-a_1$, so $\mid a_{i+1}-a_i \mid$ is constant.
Then $\mid R(a_i)-R(a_j)\mid = \mid a_i-a_j \mid, \forall i, j$
Since $\mid a_{i+1}-a_i \mid = c, \forall i$ which will be true when orbit = $1$ or $2$,
so $R(R(k))=k$ or $R(k)=k$, we can conclude that $R(R(k))=k$
Please suggest how to proceed. Thank you.
Unless there is something I don't understand, the assertion is false. An easy example is $F(x)=1$ and $G(x)=1-x$. Then $R(0)=1$, $R(1)=\infty$ and $R(\infty)=0$. So for $n=3$, we have $Q(0)=R^3(0)=0$, but $R^2(0)\ne 0$.