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 $P(s)=s$ is the only non-constant polynomial satisfying the conditions.

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.)

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 $P(1)=1$.

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 $P(s)=s$ is the only non-constant polynomial satisfying the conditions.