Consider $x \in X$ as a representative of its orbit $O_x$. Because the translation groupoid is equivalent to its skeleton, we must have \(\Hom_{\cat{sk}\cT_GX}(O_x,O_x)\cong\Hom_{\cT_GX}(x,x) =: G_x,\) the set of automorphisms of $x$. This group consists of precisely those $g \in G$ so that $g \cdot x = x$. In other words, the group $\Hom_{\cT_GX}(x,x)$ is the stabilizer $G_x$ of $x$ with respect to the $G$-action. Note that this argument implies that any pair of elements in the same orbit must have isomorphic stabilizers. As is always the case for a skeletal groupoid, there are no morphisms between distinct objects. In summary, the skeleton of the translation groupoid, as a category, is the disjoint union of the stabilizer groups, indexed by the orbits of the action of $G$ on $X$.