Smoothness of $X$ is not needed (neither for the comparison isomorphism nor for the result in question). Let $X$ be any quasi-separated scheme 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).

[The hypothesis on $n$ is necessary because if $n = p = {\rm{char}}(k)>0$ and $X = {\rm{Spec}}(A)$ is affine then the effect of $G(k)$ on ${\rm{H}}^1(X, \mathbf{Z}/(p)) = A/\wp(A)$ (with $\wp(f) = f^p-f$) is the induced action from the $G(k)$-action on $A = \Gamma(X,O_X)$, and this is generally nontrivial (e.g., $X = G = \mathbf{A}^1_k$ with the translation action corresponding to $c.f(t) = f(t+c)$ on global functions (for $c \in G(k)$).]

By a spectral sequence argument using a covering by quasi-compact $G$-stable open subsets we may reduce to the case when $X$ is quasi-compact (and quasi-separated).
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.