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 $s>1$ ?



Does there exist a rational polynomial $P(x)\in{\mathbb Q}[x]{}$ such that $P(\zeta(s))=\zeta(P(s))$?
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 $0<\zeta(2)-1<1$).


