How to understand the structure of the interesting graph obtained from the group? Let $G = A_5$ and $H < G$ is subgroup of $A_5$ generated by $(12)(34)$, and $(125)$. 
Define graph $\Gamma$ by a vertex set is element of $G$ and elements $x$ and $y$ 
adjacent if $|H^x \cap H^y| = 1$. We obtain a regular graph with valency $36$. 
I computed with SAGE some other properties of this graph:
Order of ${\rm Aut}(\Gamma)$ is $4492687491149384908800000000000 = 2^{43} \cdot 3^{21}\cdot 5^{11}$ and act vertex, edge and arc transitive. 
${\rm Aut}(\Gamma)$ has a trivial solvable radical and 
$|{\rm Soc}({\rm Aut}(\Gamma))| = 2^{30} \cdot 3^{20} \cdot 5^{10}$, hence 
${\rm Soc}({\rm Aut}(\Gamma)) = A_6^{10}$
$\Gamma$ has diameter $2$ and spectrum $\{36^1, 6^4, 0^{50}, -12^5\}$.
Let $u, w \in \Gamma$, then
$$
|\Gamma(u) \cap \Gamma(w)| = 
\begin{cases}
 18 & \text{if u and v adjacent }  \\
 36 & \text{if u and v non adjacent and } \Gamma(u)=\Gamma(v) \\
    24 & \text{if u and v non adjacent and } \Gamma(u) \ne \Gamma(v)\\
\end{cases}
$$ 
Hence if $M$ djacency matrix of the graph, then
$M^2 = 18M+24A+36B+36I$, where $A, B$ – $01$-matrix and $M+A+B+I=J$ ($I$ - identity matrix and $J$ - all one matrix)
Let $D=\Gamma(u)$, is local graph, then
${\rm Spec}(D) = \{18^1, 6^1, 0^{32}, -12^2\}$
and $\bar D =  C_6 ⊠ K_6$ (strong product)
Can you say something more about this graph? I saw a question about a similar graph, but only his automorphism group was discussed there.
 A: Without computing it explicitly, given some of the data that you've given, this is almost certainly the lexicographic product of the complement of the Petersen graph $\overline{P}$, with an edgeless graph on $6$ vertices. 
(In other words, you take $\overline{P}$, each vertex becomes $6$ vertices, and every edge a $K_{6,6}$.) 
You could build this with Sage and check for isomorphism.
Here is my train of thought: since some vertices have the same open neighbourhood and the graph is vertex-transitive, it must be a lexicographic product with an edgeless graph. Given the socle, I guessed there are $10$ blocks of size $6$. 
The quotient then has order $10$ and valency $6$. It's automorphism group has order the one you started with, divided by $(6!)^{10}$, which is $120$. The complement is $3$-valent with automorphism group of order $120$. In particular, it is connected, and since $3$ divides the order, must be arc-transitive, and the only possibility is $P$.
In particular, this means the group is isomorphic to $(S_6)^{10}\rtimes S_5$ where  $S_5$ acts as on unordered pairs.
