How to understand the automorphism group of a very symmetric graph (related to sylow intersections) For a group $G$ and subgroup $H$, consider the relation on $G$ defined $x \sim y$ if $H^x \cap H^y = 1$. This defines a graph on $G$.
It is always fairly symmetric: $N_G(H)$ acts on the left and $G$ on the right as graph automorphisms.
For some choices of $H\leq G$ the automorphism group of the graph is much (much) larger.
For $H$ a Sylow $2$-subgroup of $G=S_5$, the resulting graph has 3840 undirected edges on 120 vertices and is extremely symmetric: the automorphism group $X$ of the graph has structure $2.A_6.2.2^4.2.2^4.2.2^4.A_8^{15}$ and acts vertex, edge, and arc-transitively. It has two orbits of non-edges. It is highly connected, each pair of vertices has at least 32 common neighbors (and in the complement graph, each pair of vertices has at least 22 common neighbors). Neither it nor its complement is bipartite.
While working on graphs defined by slightly more complicated (and irrelevant) means, I was surprised to find that nearly all of the graphs were disjoint unions of complete graphs. This is the first example that is not in some sense “fully symmetric”, but I am a bit overwhelmed by how to study it.

How does $A_8^{15}$ act on $S_5$?

$|G|=8\cdot 15$ and $|H|=8$ but I'm not sure.

Does $X$ have a normal subgroup isomorphic $S_8^{15}$?

This would account for the middle 2s, but then what is up with them grouping into singles and quadruples, but the $A_8$ grouping into a 15-tuple?

What is the socle of the quotient of $X$ by $A_8^{15}$?

Is that $S_6$ acting on 3 copies of $2^4$? Are they all the same $2^4$ or are some conjugate under the outer automorphism of S6?
 A: Any permutation of any of the cosets $Hx$ is an automorphism of the graph. So you can immediately see the subgroup $N = S_8^{15}$. I haven't attempted to prove it, but it seems clear that these cosets must form a system of imprimitivity for $X$, in which case the subgroup $N$ is indeed normal, so $X = S_{15} \wr P$, where $P$ is a transitive permutation group of degree 15 and, since it has $A_6$ as composition factor and order 720, must be $S_6$. (You can verify that from the transitive groups database.)
A: In progress:
Let $$X(G,H) = \left\{ f \in \newcommand{\Sym}{\operatorname{Sym}}\Sym(G) : H^x \cap H^y \neq 1 \iff H^{f(x)} \cap H^{f(y)} \neq 1 \right\}$$ be the group of graph automorphisms of the graph defined in the question.
Note that $\Sym\left({N_G(H)x}\right) \leq X(G,H)$ for every $x \in G$. This is simply because if $f \in \Sym\left({N_G(H)x}\right)$, then $H^{f(x)} = H^{nx} = H^x$.
If $T$ is a right transversal of $N_G(H)$ in $G$, then we get that the subgroup generated by the various $\Sym\left({N_G(H)x}\right)$ for $x \in T$ is a direct product and a subgroup of $X(G,H)$.
In general, it need not be a normal subgroup (take $G$ to be $S_4$ and $H$ to be a Sylow 2-subgroup, then $X(G,H)=\Sym(G)$ and $\Sym(8) \times \Sym(8) \times \Sym(8) \leq \Sym(24)$, but it is not normal).
If the embedding of $H$ in $G$ is “sufficiently rich” so that one can determine if $H^x = H$ based on the adjacencies of $x$, then $N_G(H)$ will generate a block system, and we will get a subgroup of a wreath product.
Now what permutes those blocks?
Why would we get the full wreath product?
