Let $X$ be $k$ variety or more genrally a $k$-scheme. Denote the algebraic closure of $k$ by $\overline{k}$. it's a fact that $X(\overline{k}):=Hom(\operatorname{Spec} \ \overline{k}, X)$ as set is dense in the underlying topological space of $X$.

Obviously the Galois group $Gal(\overline{k}/k)$ acts on $X(\overline{k})$ by composition: let $\sigma \in Gal(\overline{k}/k)$ then $\alpha: X(\overline{k}) \to X(\overline{k}), \alpha \mapsto \alpha \circ \operatorname{Spec} \ \sigma$ where $\alpha \in X(\overline{k})$ and $\operatorname{Spec} \ \sigma$ is the spec morphism induced by $\alpha$.

Question: Is it true and if yes then why that the action of $\alpha$ as described above on $\overline{k}$-valued points $X(\overline{k})$ extends in appropriate way to an action on the whole scheme $k$-scheme $X$. By density $X(\overline{k})$ of it's clear that if such extension of action exists it's unique. That is the question is if such extension always or under "weak" assumptions on $X$ always exist and how does it look like. Is there a concrete description of it?

If we consider $X$ as contravariant functor $X: (Sch/k) \to (Set), Y \mapsto Hom_k(Y,X)=X(Y)$ then obviously it suffice to show that the action of $Gal(\overline{k}/k)$ on $X(\overline{k})$ extends to an action on $X(Y)$ for every $k$-scheme $Y$. That's also not obvious.

Does somebody know when under extra assumptions on $X$ such Galois action on $X(\overline{k})$ as described above extends to $X$? My hope is that the density condition could "somehow" allow a continuation (in what sense ever) on the action.