How to define the base extension of a group action on a scheme 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).
 A: 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 $G$.
A: First, you forgot a "prime" in the $G\times_S X$ just before your last sentence. 
However the one you want to prove is a general property of fibered products. One way is to draw the diagram where you put in a "tower" your two fiber products $G\times_SX$ and $(G\times_SX)\times_XX'$ (sorry I am not able to draw here). This will give you a morphism $G\times_SX'\to (G\times_SX)\times_XX'$ and you can prove it is an isomorphism.
Note that if everything is affine your claim follows easily because you can write
\begin{equation}
(M\otimes_BN)\otimes_AP\simeq M\otimes_B(N\otimes_AP)
\end{equation}
whenever you have a ring homomorphism $B\to A$ and $N,P$ are $A$-modules, and $M$ is a $B$-module (and you are exactly in this situation on the level of morphisms of schemes).
A: edit: As Li Zhan points out, I misunderstood the question. I will leave what I had written here anyways.
Your last sentence should read "However, I don't know how to show this fibre product is $G \times_S X'$", i.e. it should be $X'$ instead of $X$.
The answer to your question follows from a more general fact, namely that towers of fibered squares are fibered squares (this is exercise 2.3.P in Vakil's FOAG, version of May, 16th).
Suppose that we have some category and two fibered squares

$W \longrightarrow Y$
$\downarrow    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,    \downarrow$
$X \longrightarrow\,\, Z$

and

$W \longrightarrow Y$
$\downarrow    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,    \downarrow$
$X \longrightarrow\,\, Z$

i.e. $W = Y\times_Z X$ and $U = W\times_X V$. Then if we put them toghether we get the diagram

$U \longrightarrow W \longrightarrow Y$
$\downarrow    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,    \downarrow$
$V \longrightarrow X \longrightarrow\,\, Z$.

The claim is now that the outer rectangle

$U \longrightarrow Y$
$\downarrow    \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,    \downarrow$
$V \longrightarrow\,\, Z$

is again a fibered square (so the horizontal morphisms are the compositions of the morphisms in the single diagrams), i.e. $U = Y\times_Z V$.
