I want to understand the proof of Theorem 1.58 in "Groups, Graphs and Trees" by John Meier. I have some problems with the second line (see screenshot).
With every connected simple graph $\Gamma$ we can associate a 1-complex $X_\Gamma$. An automorphism of $\Gamma$ is an autohomeomorphism of $X_\Gamma$ which maps vertices to vertices. Let $\sigma\colon G \to \mathsf{Aut}(\Gamma)$ be an action of a group $G$ on a graph $\Gamma$. A closed subset $\mathcal F \subset X_\Gamma$ is called fundamental domain if the set $\{\sigma(g)(\mathcal F)\mid g \in G\}$ covers $X_\Gamma$ and the subset $\mathcal F$ is minimal closed subset with this property.
Lemma. Assume that if $\sigma(g)(\mathcal F) = \mathcal F$ for $g \in G$ then $g = 1$. Then there is a point $x \in \mathcal F$ that is moved freely under the action $\sigma$ (that is, $\sigma(g)(x) = \sigma(h)(x) \Rightarrow g = h$ for any $g, h \in G$).
I can't figure out why would this statement be true. In this book there is no explanation (perhaps, author assumes that this is obvious and maybe it really is).
I would very appreciate any help!
