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.
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.