A question on "unlabeled Cayley graphs" Suppose $G$ is a finite group $S \subset G$. Let's define $UCG(G, S)$ (unlabeled Cayley graph) as a finite unordered simple graph $\Gamma(V, E)$, where $V = G$ and $E = \{(x, y) \in G \times G| x \neq y \text{ and } (y^{-1}x \in S \text{ or } x^{-1}y \in S)\}$. 
Unlabelled Cayley graph satisfy following properties:

$UCG(G, S)$ has $[G: \langle S \rangle]$ components and each of them is isomorphic to $UCG(\langle S \rangle, S)$

Trivially follows from the definition

A finite simple undirected graph $\Gamma$ is isomorphic to $UCG(G, S)$ for some finite group $G$ and its subset $S$ iff there exists such a group $H$ and its subset $X$, that each component of $\Gamma$ is isomorphic to $UCG(H, X)$

=> Trivially follows from the first proposition
<= Suppose $\Gamma$ has $n$ components. Then we can take $G = H \times C_n$ and $S = X \times \{e\}$.

$UCG(G, S)$ is vertex transitive

Consider the action of $G$ on itself by left multiplication.

$G \leq Aut(UCG(G, S))$

Consider the action of $G$ on itself by left multiplication.
My question, however, is:

Does for any finite simple undirected vertex transitive graph $\Gamma$, there exist such a finite group $G$ and its subset $S$, that $\Gamma \cong UCG(G, S)$?

Because of the second proposition we can without loss of generality also assume, that $\Gamma$ is connected. However, that did not help me much...

According to Lauri, Josef; Scapellato, Raffaele (2003), Topics in graph automorphisms and reconstruction, the Petersen graph is not a Cayley graph of any group $G$ and generating set $S$. There are only two groups of order $10$, $\mathbb Z/10\mathbb Z$ and $D_5$. For each, you can look at all generating sets for which $|S\cup S^{-1}|=3$ (necessary because the Petersen graph is $3$-regular), and note that all of them would result in a graph with diameter more than $2$, whereas the Petersen graph has diameter $2$.
