How does an automorphism of vertices stabilize edges? How an automorphism of vertices stabilizes edges ?
There are some permutations which acts on vertices and edges at the same time. For example, 
$\pi=(24)$(an automorphism)  permutes or switches vertex $2,4$ also moves edges $a,b,c,d$($a$ switches with $b, c$ switches the place with $d$) and  fixes(stabilizes)  vertex $5$ and edge $e$.
But for $X \setminus\{ 5 \}$, which is a cyclic graph $C_4$, there is no automorphism that fixes(stabilizes) any edge(right?). If that is the case, then how an automorphism of vertices stabilizes edges ?
Context :-
From Lecture 2, Algebra and Computation by V. Arvind, (page3), it is given -

The Problem Statement: $H \leq  S_n$, given by a small generating set.
  Also given is a subset $\Delta \subset [n]$. Find the stabilizer of $\Delta$ in $H$,
  defined as $stab_\Delta(H) = \{\pi \in H : \Delta^{\pi} = \Delta \}$ Though this >problem has
  nothing to do with graphs directly, graph isomorphism reduces to this
  problem (which we shall call Set-Stab).
Theorem 3.
  Graph-Iso $\leq_P$ Set-Stab
Proof. By our earlier theorem, it is enough to show that Graph-Aut
  reduces to Set-Stab. One simply needs to note that an automorphism can
  be thought of as acting on the edges as well. Given a graph 
  $G = (V,E)$, a permutation of the vertices induces a permutation of the edges.
  Hence,
$\phi : Sym(v) \mapsto Sym({{V}\choose{2}})$ is injective.
Thus, all we need to do is find the set of elements in $Sym(V )$ that
  stabilizes $E$. Taking the set H to be 
  $ \phi (Sym(V )) \subset Sym({{V}\choose{2}})$ and  $\Delta = E \subset [{{V}\choose{2}}]$,  the
  automorphism group is precisely $stab_{\Delta}(H)$.

Also $H \leq Sym({{V}\choose{2}})$ which has  a "relation" with  $Sym(v) $ (according to $\phi$ ). This relationship is "injective", i.e. for each element of $H$ there is  only one element in  $Sym(v)$. But, $\phi$  is not "surjective (onto)" .
How the automorphism group is precisely $stab_{\Delta}(H)$?
I have asked  a question  on the same lecture note which might be relevant to mention . Note that the proof started by saying "By our earlier theorem,", I asked about theorem in that question.
Edit: $\Delta$ is a subset of vertex set= $[n]$.
 A: For a simple graph, an automorphism is defined as a permutation of the vertices that preserves adjacency; this induces a permutation of the edges.  So in the example you give, there is exactly one nontrivial automorphism of the graph given by $(2 \ 4)$, it induces the permutation $(a \ b)(c \ d)$ on edges so the edge labelled $e$ is stabilized. 
If you remove vertex $5$, you have a graph $C_{4}$ with automorphism group $G \simeq D_{4}$ (the dihedral group, which has order 8).  You do have nontrivial automorphisms stabilizing edges; for example, the map $(1\ 2)(3 \ 4)$ induces the permutation $(b\ c)$ on edges and stabilizes both $a$ and $d$.
The automorphism group $\mathrm{Aut}(X) \leq \mathrm{Sym}(v)$ and is also isomorphic to some subgroup of $\mathrm{Sym}( {v \choose 2})$; the best way to view ${v \choose 2}$ in this context is as the collection of 2-element subsets from $[v]$, then instead of $\mathrm{Sym}({v \choose 2})$ you can just work in $\mathrm{Sym}(v)$ and consider the action it induces on the 2-element subsets.  The edges $E$ of the graph should also be viewed as 2-sets of vertices, so that $E \subseteq {v \choose 2}$; this set $E$ is what they mean by $\Delta$. 
A permutation $\tau \in \mathrm{Sym}(v)$ is in $\mathrm{Aut}(X)$ if and only if it preserves adjacencies, that is, if and only if it stabilizes the set of edges; this is what the mean by $\tau \in \mathrm{Stab}_{\Delta}(H)$. Note that they take $H = \phi(\mathrm{Sym}(v)) \subset \mathrm{Sym}({v \choose 2})$; if you view ${v \choose 2}$ as the collection of $2$-sets as I mentioned, you can think of $H = \mathrm{Sym}(v)$.
