Let $G$ be an algebraic group (not necessarily linear) defined over an
algebraically closed field $k$,
acting on a smooth integral $k$-variety $X$.
Let $x_0\in X(k)$ and let $\pi_1(X,x_0)$ denote the étale
(Grothendieck's) fundamental group of $X$.
Assume that either $G$ fixes $x_0$ or the group $\pi_1(X,x_0)$ is abelian.
In both cases $G(k)$ acts on $\pi_1(X,x_0)$.
I need a proof that if $G$ is *connected*, then this action is trivial.

I know a proof in characteristic 0. In this case by the Lefschetz principle we may assume that $k=\mathbf{C}$, and we can consider the action of $G(\mathbf{C})$ on the topological fundamental group $\pi_1^{\mathrm{top}}(X(\mathbf{C}),x_0)$. Let $g\in G(\mathbf{C})$. Since $G$ is connected, we can connect $g$ with the unit element $e\in G(\mathbf{C})$ by a continuous path. We see that the automorphism $g_*\colon X(\mathbf{C})\to X(\mathbf{C})$ is homotopic to the identity automorphism. It follows that the induced automorphism of $\pi_1^{\mathrm{top}}(X(\mathbf{C}),x_0)$ is the identity.

I would like to see a proof that the action is trivial in *arbitrary*
characteristic.