I don't know how to state my question precisely in the language of group schemes. You have an algebraic group $G$, defined over a perfect field $k$, and the Galois group $\Gamma = \textrm{Gal}(\overline{k}/k)$ acts on $G$. The $k$-rational points $G(k)$ consist of those $x \in G$ which are fixed by $\Gamma$.

For fixed $\sigma \in \Gamma$, $x \mapsto x^{\sigma}$ is a bijection of $G$, but not an isomorphism of varieties.

My questions:

1 . Is this mapping $x \mapsto x^{\sigma}$ in general an abstract group isomorphism?

2 . How do you state these results in the language of group schemes?

An example, consider $G = U(2,1)$ as a group over $\mathbb{R}$. As a set, $G = \textrm{GL}_3(\mathbb{C}$). If $\sigma \in \Gamma = \textrm{Gal}(\mathbb{C}/\mathbb{R})$ is complex conjugation, then $\Gamma$ acts on $G$ as

$$x^{\sigma} = w \, ^t \bar{x}^{-1}w$$

where $w = \begin{pmatrix} & & 1 \\ & -1 & \\ 1 & & \end{pmatrix}$ and $\bar{x}$ denotes entrywise complex conjugation. This is a group homomorphism.