TITLE: How to define the base extension of a group action on a scheme
QUESTION [6 upvotes]: Suppose $G/S$ is a group scheme over $S$, $X/S$ is a scheme over $S$. $G$ acts on $X$ by the morphism $ \sigma : G \times_S X \to X$. Let $X'$ be a scheme over $X$. How to deine the group action on $X'$, i.e. $\sigma ' : G \times_S X' \to X'$ ?
Presumabely, this morphism should be the base extension of $\sigma$, i.e. $(G \times_S X) \times_X X'$. However, I don't know how to show this fibre product is $G \times_S X'$. (WLOG, one can assume $X,X'$ are affine schemes).

REPLY [3 votes]: You can always make $G$ act trivially on $X'$. But presumbly you want the action be compatible with the action on $X$. Then it is not possible in general: just take for $X'$ a closed point of $X$ not stable by the action of $G$.