How to understand the automorphism group of a very symmetric graph (related to sylow intersections)

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.)