If $\Gamma = C(G, S)$ is the (undirected) Cayley graph of a finite group $G$ with generating set $S$, then $G \le \operatorname{Aut}(\Gamma)$, the "full" automorphism group of $\Gamma$. 



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> When is it true that $G \cong \operatorname{Aut}(\Gamma)$? In other words, when is a group $G$ (isomorphic to) the automorphism group of its own Cayley graph (via a specified generating set)?
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Here, by "full" automorphism group, I mean the group of permutations of the vertex set that preserve adjacency (with no regard to the coloring that is usually associated with Cayley graphs).



Condition(s) for the full autormophism group $\operatorname{Aut}(C(G, S))$ of the Cayley graph of $G$ to be isomorphic to $G$
First, some terminology: if $G\cong\mathrm{Aut}(\Gamma)$ then $\Gamma$ is often called a GRR (for graphical regular representation). This may help in looking for references. Determining whether a Cayley graph is a GRR given $G$ and $S$ is very difficult in general.


One necessary criterion is the following: let $\mathrm{Aut}(G,S)$ be the group of automorphisms of $G$ that preserve $S$. If $\Gamma$ is a GRR then $\mathrm{Aut}(G,S)=1$. 


Sometimes, this is also sufficient, see for example:


-Godsil, C. D.; On the full automorphism group of a graph. Combinatorica 1 (1981), 243–256. 


-Li, Cai Heng; Sim, Hyo-Seob; The graphical regular representations of finite metacyclic p-groups. European J. Combin. 21 (2000), 917–925.


In fact, the semidirect product of $G$ and $\mathrm{Aut}(G,S)$ is exactly the normaliser of $G$ in $\mathrm{Aut}(\Gamma)$, so that part is somewhat under control, but the part of $\mathrm{Aut}(\Gamma)$ that does not normalise $G$ is very difficult to deal with in general.


