Does there exist a rational polynomial $P(x)\in{\mathbb Q}[x]{}$ such that $P(\zeta(s))=\zeta(P(s))$? let $P(x)\in{\mathbb Q}[x]{}$ be a rational polynomial with $P(1) >1$ and $\zeta $ be the Riemann zeta function , I want to know if there exist a rational polynomial such that $P(\zeta(s))=\zeta(P(s))$ with $s>1$ ?
 A: No. If $P$ has a positive leading coefficient, then letting $s$ go to infinity and using continuity of $P$ we get $P(1)=1$. If $P$ has a negative leading coefficient, then $\zeta(P(s))$ has zeros at arbitrarily large $s$, while $P(\zeta(s))$ does not.
A: It is straightforward to see that if $P\in\mathbb{C}[x]$ satisfies $P(\zeta(s))=\zeta(P(s))$, then $P(1)=1$. Note that if $P(\zeta(s))=\zeta(P(s))$ holds for all real $s>1$, then it also holds for all complex $s\neq 1$ by the uniqueness of analytic continuation.
Indeed, $\zeta(s)$ has a pole at $s=1$, hence $P(\zeta(s))=\zeta(P(s))$ also has a pole at $s=1$. This implies that $P(1)=1$.
Added. It seems easy to show that $P(x)=x$ is the only non-constant polynomial satisfying the conditions. (Indeed, this is true, see Fedor Petrov's response.)
A: Here is a simple argument which applies to polynomials in $C[x]$, and even to most rational functions in $C(x)$. $\zeta$ is a meromorphic function in the plane.
So the Nevanlinna characteristic $T(r,\zeta)$ is defined. It is a positive increasing function, a kind of transcendental analog of the degree of a rational function. Your equation implies
that $(\deg P+o(1))T(r,\zeta)=T(r^d(1+o(1),\zeta).$ Here $d$ is the order of pole
of $P$ at $\infty$, for the case of polynomial, $d=\deg P$, of course.
If $d\leq 0$ then the right hand side is bounded, if $d\geq 1$, we obtain a contradiction since for every transcendental meromorphic function $T(r,f)/\log r\to\infty$. So only the case $d=1$ remains, but for this case the completion of the proof is elementary. Of course I understand that this is too heavy artillery to solve such an elementary question:-)
A: Extending the argument by GH from MO, $\zeta(P(s))$ has a pole for any $s$ such that $P(s)=1$, while $P(\zeta(s))$ has unique pole for $s=1$. Therefore if $\zeta(P(s))=P(\zeta(s))$, then $P(s)=1$ has unique solution $s=1$ and $P(x)-1=c(x-1)^n$ for some complex $c$ and positive integer $n$, and we get $\zeta(1+c(s-1)^n)=1+c(\zeta(s)-1)^n$. For $s=1+x$ with small $x$ the equality of leading asymptotic terms gives $1/(cx^n)=c/x^n$, $c=\pm 1$. For $s=2$ we get $\zeta(1+c)=1+c(\zeta(2)-1)^n$, i.e., either 
(i) $c=1$, $\zeta(2)-1=(\zeta(2)-1)^n$, $n=1$, $P(x)=x$, or
(ii)$c=-1$, $-3/2=-(\zeta(2)-1)^n$, this is impossible (since $0<\zeta(2)-1<1$).
