In process of solving a problem, I had to find some ordered pair of polynomial $(f(x),g(x))$ with coefficient $\mathbb{Z}_{p}$, $p$ is prime number. (i.e $f(x),g(x) \in \mathbb{Z}_{p}[x])$ such that $\frac{f(x+r)}{g(x+r)}=\frac{f(x)}{g(x)}$ for all $r \in \mathbb{Z}_{p}$.
I find out that $\mathbb{Z}_{p} \simeq \{(r,1) \mid r\in \mathbb{Z}_{p} \}$ is one of them, but can't find out another. i thought about $f(x^{p}), g(x^{p})$ because $(x+r)^p=x^{p}+r^p$ but $r^p$ doesn't disappear when $1 \leq r <p $ by Fermat's theorem. is there something else? and if there's no more, how can i prove it?
P.s.: $x$ is transcendental over $\mathbb{Z}_{p}$, so all the members of $\left\{\frac{f(x)}{g(x)} \ \middle\vert\ f(x),g(x) \in \mathbb{Z_p} \right\}$ is simple extension of $x$ over $\mathbb{Z_p}$.
You are looking for the fixed points of the group of automorphisms $G$ of the function field $L=\Bbb{Z}_p(x)$, where $G$ consists of the automorphisms $$ \sigma_r:\frac{f(x)}{g(x)}\mapsto\frac{f(x+r)}{g(x+r)} $$ and $r$ ranges over $\Bbb{Z}_p$. We have $\sigma_{r_1}\circ\sigma_{r_2}=\sigma_{r_1+r_2}$, so as a group it is isomorphic to the additive group of $\Bbb{Z}_p$. In particular, it is cyclic of order $p$.
By general Galois theory it follows that the fixed field of $G$ is a subfield $K=\operatorname{Inv}(G)$ of $L$ such that $[L:K]=p$. We need to identify $K$.
Your idea of using the Frobenius is good. All that is missing is to recall Little Fermat, stating that $r^p=r$ for all $r\in\Bbb{Z}_p$. It follows that $u=x^p-x\in K$. Simply because $$ \sigma_r(x^p-x)=(x+r)^p-(x+r)=x^p+r^p-x-r=x^p-x. $$ It follows that the purely transcendental extension $F=\Bbb{Z}_p(u)\subseteq K$. I claim that we actually have equality here, $K=F$. This follows from the observation that the relation $$ x^p-x=u $$ implies that $x$ is algebraic of degree $\le p$ over $F$. This, in turn, implies that $[L:F]\le p$. Together with the Galois theoretic fact the claim follows.