$\textbf{Theorem}$ - A group $G$ acts doubly transitively on a set $X$ iff $1/|G|\sum_{g\in G}|fix(g)|^2=2$.
I Have no idea how to begin. If it had been finite group and finite set, then at the most I could say was $|G|=\sum_{g\in G}|fix(g)|$ by Cauchy-Frobenius Lemma/Burnside Lemma. But sets and group here can be infinite too. And what role double transitivity is playing? I guess rank of double transitive group is $2$ must be used , but not sure how to get $|fix(g)|^2$ into the picture
If we note $X_g$ for $fix(g)$ we have for the Cauchy-Frobenius Lemma that $\sum_{g\in G}|X_g|$ equals $|G|$ times the number of orbits in $X$. This is easily proved by the following argument: Let $x\in X$ then $x\in X_g \iff g\in S_x$ where $S_x \subseteq G $ is the stabilizer group of $x$. The number of times $x$ occurs in one of the $X_g$ equals the number of times $g \in S_x$, and this is $|S_x|$. If we group the elements of $X$ by orbit then we have $\sum_{g\in G}|X_g|=\sum_{O\in\mathcal{O}}|O||S_{\bar{O}}|=\sum_{O\in\mathcal{O}}|G|$ by the orbit-stabilizer lemma, where $\mathcal{O}$ is the set of orbits $O$, and $\bar{O}$ is a representative element of the orbit $O$. We follow the same reasoning for $\sum_{g\in G}|X_g^2|$. We have that $(x,y)\in X_g^2 \iff a,b\in X_g \iff g\in S_{(a,b)}$ so we can apply the the Cauchy-Frobenius Lemma to the space $X\times X$ which gives $2|G|$ since $X\times X$ has two orbits iff the action of $G$ on $X$ is doubly transitive.