Let $G$ be a group scheme over a basis $S$, and $X$ be a scheme over $S$. Let $\rho: G \times_SX \to X$ be an action of $G$ on $S$. If $T$ is an $S$-scheme and $x \in X(T)$, then the definition of the stabilizer of $x$ is given as the subgroup scheme of $G_T$ that represents the functor $T' \mapsto \{g \in G(T') \mid g \cdot x = x\}$ on $T$-schemes $T'$.
I have problem understanding this definition. Should the $G(T')$ actually be a $G_T(T')$, since we're defining a subgroup scheme of $G_T$? And how does $g \in G(T')$ (or $G_T(T')$) act on $x \in X(T)$? We only know a priori that $G(T')$ acts on $X(T')$.
$G(T')$ and $G_T(T')$ can be naturally identified: if $p: G_T = G \times_S T \rightarrow G$ is projection onto $G$, then $h \mapsto p \circ h$ defines a bijection
$$G_T(T') = \textrm{Hom}_T(T', G \times_S T) \rightarrow \textrm{Hom}_S(T',G) = G(T')$$
$G(T') = G_T(T')$ does not act on $X(T)$, but the structure morphism $f: T' \rightarrow T$ induces a map $f_{\ast}: X(T) \rightarrow X(T')$. To be precise, they should say that the $T'$-points of the stabilizer of $x$ consist of those $g \in G(T')$ which fix $f_{\ast}(x)$.