I'm reading this part of The Stacks Project regarding ramification theory. Right before defining decomposition and inertia groups for schemes there is the following passage:
Let $X$ be a normal integral scheme with function field $K$. Let $K^\text{sep}$ be a separable algebraic closure of $K$. Let $X^\text{sep}\rightarrow X$ be the normalization of X in $K^\text{sep}$. Since $G=\operatorname{Gal}(K^\text{sep}/K)$ acts on $K^\text{sep}$ we obtain a right action of G on $X^\text{sep}$.
It is not clear to me why and how $G$ acts on $X^\text{sep}$. Little hint?
Saying that $X^{sep}$ is the normalization means that affine-locally on $U=\operatorname{Spec} A\subset X$ the map $X^{sep}\to X$ corresponds to the natural inclusion from $A$ to it's integral closure $\overline{A}\subset K^{sep}$. As $Gal(K^{sep}/K)$ acts on $\overline{A}$ preserving the inclusion $A\to \overline{A}$, we see that $Gal(K^{sep}/K)$ acts by automorphisms on the map $U^{sep}\to U$. As normalization commutes with localization, we may patch these actions on open affines together to see that $Gal(K^{sep}/K)$ acts via automorphisms on $X^{sep}\to X$.