Is there any relation between automorphism group of a Cayley graph over a group and over its subgroup? Let $\Gamma=Cay(G,S)$ be a Cayley graph over a group $G$, $H$ be a proper subgroup of $G$ and $\Sigma=Cay(H,T)$ where $S$ and $T$ are inversed-closed subsets of $G$ and $H$ not containing idendity, respectively. Is there any relation between $Aut(\Gamma)$ and $Aut(\Sigma)$ in general? When $T=H\cap S$, what can we say?
 A: I am reasonably confident that the real answer to your question is "No". For example,
if $T=H\setminus 1$ then the automorphism group of $\mathrm{Cay}(H,T)$ is the full
symmetric group. For another, take $S=T$; then $\Gamma$ will have a large automorphism
graph, a wreath product. (If you want your Cayley graphs connected, take complements.)
So adding generators may reduce symmetry, or increase it.
Note that questions phrased as you have are a bit difficult to answer, since the
answer depends on what you mean by "any relation between".
A: I don't know how interested you are in the kinds of things that happen in infinite groups, but this question does have some general interest in that setting. 
It seems to be quite common, for example, that $G$ has finite index in the automorphism group of its Cayley graph. This is true, for example, for many examples of lattices in Lie groups, as Alex Furman has noted. For a very specific example, this is true for the fundamental group of a closed surface of genus $g$ with the standard presentation
$$\langle a_1,b_1,\ldots,a_g b_g \quad | \quad [a_1,b_1] \ldots [a_g,b_g] \rangle
$$
At a different extreme, for free groups with their standard generating set, the automorphism group of the Cayley graph is a locally compact topological group locally homeomorphic to the Cantor set, and hence the free group has uncountable index. 
