I would like to enumerate Cayley graphs ${\rm Cay}(G,C)$ on $n$ vertices. Since we have to choose a subset $C$ with $m$ elements out of the group $G$ (excluding identity) then there are ${n-1 \choose m}$ such subsets, where $m$ must be even because for every $x \in C$ we have $x^{-1} \in C$ (assuming that no element is its own inverse). So there must be $\sum_{l=1}^{\frac{n-1}{2}} {n-1 \choose 2l}$ such graphs, right?
And if there are elements which are their own inverse (say, $k$ of them) then would we have $\sum_{l=1}^{\frac{n-k-1}{2}} {n-k-1 \choose 2l} + k$ such graphs?
Suppose your group $G$ is fixed, and it has $n$ elements, $k$ of which are self-inverse. You want to count the number of ways to pick a self-inverse subset $C\subseteq G$. You can pick the self-inverse elements as you wish, so that's $2^k$ choices. The non-self-inverse elements come in $\frac{n-k}{2}$ pairs, for $2^{\frac{n-k}{2}}$ choices.
The total number of (unlabelled) Cayley graphs on $G$ is thus $$2^k2^{\frac{n-k}{2}}=2^{\frac{n+k}{2}}.$$
(Here I was assuming that we allow $1\in C$. Otherwise, make the obvious modification.)