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. 



A characteristic-free proof that the action of a connected algebraic group $G$ on the fundamental group of a $G$-variety is trivial
This is false as stated in positive characteristic. For example, suppose that the characteristic of $k$ is $p$; take $X = \mathbb A^1\_k = \mathop{\rm Spec} k[t]$ and $G = \mathbb G\_{\rm m}$. If $a \in k^\*$ and $E$ is the standard étale cover $E = \mathop{\rm Spec} k[x,t]/(x^p - x - t) \to \mathop{\rm Spec} k[t]$ of $X$, its pullback through multiplication by $a$ is $ \mathop{\rm Spec} k[x,t]/(x^p - x - at) \to \mathop{\rm Spec} k[t]$, which will not be isomorphic to $E$ in general.


However, it is true when $X$ is proper over $ \mathop{\rm Spec} k$. The point is that in this case we have $\pi\_1(X \times G, (x\_0, 1)) = \pi\_1(X, x\_0) \times \pi\_1(G, 1)$ (see, for example, Corollary 5.6.6 in Szamuely's wonderful book on the fundamental group); this uses the hypothesis that $G$ is connected. The embedding of $\pi\_1(G, 1)$ corresponding in the decomposition above is induces by the embedding $G \subseteq X \times G$ defined by $g \mapsto (x\_0, g)$. If $\alpha \colon X \times G \to X$ is the action, let us show that $\alpha\_\*\colon \pi\_1(X \times G, (x\_0, 1)) \to \pi\_1(X, x\_0)$ coincides with the homomorphism $\pi\_1(X \times G, (x\_0, 1)) \to \pi\_1(X, x\_0)$ induced by the first projection. From the formula above we see that it is sufficient to show that the composite $G \to X \times G \to X$, where the first map is $g \mapsto (x\_0, g)$ and the second is the action, induces a trivial map on fundamental groups. But the map is in fact constant, because $G$ fixes $x\_0$, so this is clear.


Now fix $g \in G(k)$. The composite $X \to X \times G \to X$, where the first map is $x \mapsto (x,g)$, and the second is the action, induces the action of $g$ on $X$. On the other hand, the composite $X \to X \times G \to X$, where the first map is $x \mapsto (x,g)$, and the second is the projection, induces the identity. From the previous fact, we have that these two maps induce the same homorphism on fundamental groups, and we are done.


