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 $s>1$ ?

1$\begingroup$ I have been working on a related problem, and my conclusion is that no injective $P$ fulfilling your requirements exists. $\endgroup$ – Sylvain JULIEN Mar 25 at 23:03

2$\begingroup$ @SylvainJULIEN: Polynomials of degree at least $2$ are not injective on $\mathbb{C}$. $\endgroup$ – GH from MO Mar 25 at 23:06

2$\begingroup$ Indeed, but the OP doesn't forbid degree $1$ polynomials. $\endgroup$ – Sylvain JULIEN Mar 25 at 23:08

1$\begingroup$ The question doesn't match with the title. $\endgroup$ – YCor Mar 25 at 23:42
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 $P(x)1=c(x1)^n$ for some complex $c$ and positive integer $n$, and we get $\zeta(1+c(s1)^n)=1+c(\zeta(s)1)^n$. 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 $0<\zeta(2)1<1$).
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 nonconstant polynomial satisfying the conditions. (Indeed, this is true, see Fedor Petrov's response.)

$\begingroup$ Can your argument be used to show that any complex function commuting to $\zeta$ is necessarily continuous? $\endgroup$ – Sylvain JULIEN Mar 25 at 23:29

$\begingroup$ @SylvainJULIEN, nice idea probably you meant uses of Voronin's universality to show any commuting complex function to R zeta function must be continious $\endgroup$ – zeraoulia rafik Mar 25 at 23:32

$\begingroup$ @GH from MO , would be the same simple proof with s lie in the critical strip ? $\endgroup$ – zeraoulia rafik Mar 25 at 23:34

1$\begingroup$ @SylvainJULIEN: There are also plenty of continuous functions from $\mathbb{C}$ to $\mathbb{C}$ that permute the zeros of $\zeta(s)$, not just the identity and complex conjugation. $\endgroup$ – GH from MO Mar 25 at 23:52

2$\begingroup$ @SylvainJULIEN: I think there are plenty of continuous and noncontinuous functions from $\mathbb{C}$ to $\mathbb{C}$ that commute with $\zeta$. On the other hand, there probably very few holomorphic of meromorphic functions that commute with $\zeta$. At any rate, I stop here a this site is not a discussion board, and also these questions appear to be pretty random (they have little to do with $\zeta$). $\endgroup$ – GH from MO Mar 26 at 0:31
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.

$\begingroup$ Doesn't your argument also apply for polynomials in $\mathbb{R}[x]$? $\endgroup$ – Sylvain JULIEN Mar 25 at 23:11

2$\begingroup$ @SylvainJULIEN: There is a simple proof for all complex polynomials. See my post below. $\endgroup$ – GH from MO Mar 25 at 23:20
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:)