Let a Cayley graph $G$ of a group $H$ with respect to the generating set $\{s_i\}$ have a clique of order $> 2$. In addition assume the graph $G$ is non-complete. If the clique size is less than half the order of $G$, then is it possible for some group $H$ that $G$ has a unique "disjoint maximal clique". By "disjoint maximal clique", I mean a clique equal to the clique size of the graph, and such that any other clique of same order would not be vertex disjoint with the prior clique.

I don't think so. For, if $(e),(s_1),(s_1\cdot s_2),(s_1\cdot s_2\cdot s_3),\ldots,(s_1\cdot s_2\cdots s_n)$ be the sequence of vertices in a maximal clique, then I think even $(s_1^2),(s_1^3),(s_1^2\cdot s_2),\ldots,(s_1^2\cdot s_2\cdots s_n)$ would also be a sequence of vertices in a maximal clique, where $e$ denotes the identity element. But, what if $s_1$ is an order $2$ or $3$ element. How do we ensure that there always exist a disjoint clique apart from the clique $(e),(s_1),(s_1\cdot s_2),(s_1\cdot s_2\cdot s_3),\ldots,(s_1\cdot s_2\cdots s_n)$? Will this be true at least for the case when $H$ is an abelian/cyclic group? Any hints? Thanks beforehand.