Smoothness of $X$ is not needed (neither for the comparison isomorphism nor for the result in question).  Let $X$ be any scheme locally of finite type over a separably closed field $k$, equipped with an action by a connected $k$-group scheme $G$ of finite type.  Let $n > 0$ be an integer not divisible by the characteristic of $k$ and choose an integer $i \ge 0$. Then we want to show that the action of $G(k)$ on ${\rm{H}}^i(X, \mathbf{Z}/(n))$ is trivial (using etale cohomology here).  

It is harmless to make the radiciel extension from $k$ to its algebraic closure ("topological invariance" of etale cohomology), so we may assume that the separably closed $k$ is even algebraically closed. We may then replace $G$ with $G_{\rm{red}}$ so that $G$ is smooth. 

Let $f:G \times X \rightarrow G$ be the projection map.  The hypothesis on $n$ and smoothness of $G$ allow us to apply the smooth base change theorem to conclude that $\mathscr{F} = {\rm{R}}^if_{\ast}(\mathbf{Z}/(n))$ is the constant sheaf on $G$ attached to ${\rm{H}}^i(X,\mathbf{Z}/(n))$.
Consider the action automorphism 
$$\alpha: G \times X \simeq G \times X$$
defined by $(g,x) \mapsto (g, gx)$. This commutes with $f$, and so induces an automorphism $[\alpha]$ of $\mathscr{F}$ on $G$. The effect on the stalk at $g \in G(k)$ is the $g$-action on ${\rm{H}}^i(X, \mathbf{Z}/(n))$ that we want to be trivial.  But $\mathscr{F}$ is a *constant* sheaf on a *connected* scheme, so the effect on $\mathscr{F}$ of any automorphism is uniquely determined by the effect on a single stalk.  Looking on the stalk at $g=1$ thereby shows that $[\alpha]$ is the identity automorphism. Now pass to the effect of $[\alpha]$ on the stalk at any $g \in G(k)$ to conclude.