Given a simple, undirected graph and a vertex $v$ of the graph, let $L\_v$ denote the set of automorphisms of the graph that fixes the vertex $v$ and each of its neighbors. When the graph is vertex-transitive, the vertex-neighborhood stabilizer $L\_v$ is independent of the choice of $v$. 


I noticed that for many vertex-transitive graphs, $L\_v$ happens to be either trivial or is isomorphic to the direct product of copies of $C\_2$. Thus $L\_v$ happens to be isomorphic to $C\_2^k$ for some $k \ge 0$. For example, for the modified bubble-sort graph on 24 vertices, $L\_v$ is the Klein four-group $C\_2 \times C\_2$, and for many Cayley graphs generated by transposition sets, $L\_v$ is trivial (so $k=0$ in this case) (cf. <http://arxiv.org/abs/1205.5199>). For the complete transposition graph, $L\_v \cong C\_2$ (cf. <http://arxiv.org/abs/1404.7363>). I also considered the Petersen graph, and again $L\_v \cong C\_2$. Is this a coincidence or is there some result that says that for some families of vertex-transitive graphs or for certain families of Cayley graphs, $L\_v \cong C\_2^k$ for some $k$? What are some counterexamples - for example, what are some (vertex-transitive) graphs for which $L\_v \cong C\_3$, say?



Structure of the stabilizer of a vertex-neighborhood of a vertex-transitive graph
Consider the Kneser graph $K(v,k)$, with vertices the $k$-subsets of $V=\{1,\ldots,v\}$,
where the $k$-subsets are adjacent if they are disjoint. Let $\alpha=\{1,\ldots,k\}$
and let $G$ be the subgroup of the symmetric group on $V$ that fixes each element
of the complement of $\alpha$. So $|G|=k!$ and $G$ is a subgroup of the stablizer of $\alpha$
that fixes each neighbour of $\alpha$ in the Kneser graph. 


To get examples as you requested, take $k\ge 3$ (and $v>2k)$.


Cayley graphs are not a good place to look, because the stabilizer of a vertex tends to consist of automorphisms of the underlying group, and any automorphism that fixes each element in a connection set is the identity.


