I feel like for this question it is just a matter of showing the mapping in both directions, from the group to the graph and the graph to the group.
So for the mapping from the group to the graph, I mapped the each group action in G to some path that a vertex will travel. The path will be composed of the edges corresponding to the generators of the group action. I then chose some arbitrary vertex and showed that mapping it in such a way is an automorphism as the ends would be the same. In these automorphisms, every vertex follows the same type of path (generator/edge sequence) to reach it's mapped vertex.
Now, for the other direction I am a little stuck. I initially tried to prove by contradiction: Suppose there exists an automorphism that does not map according to a group action. This would mean the generator/edge sequence would be different for each vertex. I then noted that there must exist one edge e such that it's end vertices will follow different generator/edge sequences to reach it's mapped destination. However, I realized that it's possible for them to still be neighbors after the mapping. I hope I haven't overlooked anything or interpreted graph automorphisms wrong!
Thanks for your help!