Prove that $G\curvearrowright X$ is 2-transitive if and only if $\frac{1}{\left|G\right|}\sum_{g\in G}\left|{\rm Fix}_{X}\left(g\right)\right|^{2}=2$ Question
Prove that $G\curvearrowright X$ is 2-transitive if and only if $\frac{1}{\left|G\right|}\sum_{g\in G}\left|{\rm Fix}_{X}\left(g\right)\right|^{2}=2$
My attempt
I tried to look at the action $G\curvearrowright X\times X$ defined by $g.\left(x_{1},x_{2}\right)=\left(g.x_{1},g.x_{2}\right)$ and I got stuck there.
I tried to use Burnside's Lemma but $G$ and $X$ not necessarily finite.
Assuming that $G$ and $X$ is finite:
Using Burnside's Lemma -
$\left|_{G}\setminus^{X\times X}\right|=\frac{1}{\left|G\right|}\sum_{g\in G}\left|Fix_{X\times X}\left(g\right)\right|\underbrace{=}_{\left(*\right)}\frac{1}{\left|G\right|}\sum_{g\in G}\left|Fix_{X}\left(g\right)\right|^{2}$
(*) I think $\left|Fix_{X\times X}\left(g\right)\right|=\left|Fix_{X}\left(g\right)\right|^{2}$ becuase : 
$\left|Fix_{X\times X}\left(g\right)\right|=|\left\{ \left(x_{1},x_{2}\right)\in X\times X\,|\,g.\left(x_{1},x_{2}\right)=\left(g.x_{1},g.x_{2}\right)=|\left(x_{1},x_{2}\right)\right\}| = \left|\left\{ \left(x_{1},x_{2}\right)\in X\times X\,|\,g.x_{1} 
 =      x_{1}\land g.x_{2}=x_{2}\right\} \right| = \left|Fix_{X}\left(g\right)\right|^{2}$ 
But I still don't know why $\left|_{G}\setminus X\right|=2$
Thanks

What you have by considering the action of $G$ on $X\times X$ is that
$$
|G\backslash X\times X| = \frac{1}{|G|}\sum_{g\in G} |\text{Fix}_X(g)|^2
$$
(Note that you have written $|G\backslash X|$ in you post which is incorrect).
Now the fact that $G$ acts doubly transitively on $X$ does not say that $G$ affords only one orbit on $X$. The definition is that whenever $x_1, x_2$ are distinct in $X$ and $y_1, y_2$ are distinct in $X$, one can find $g\in G$ such that $gx_1=y_1$ and $gx_2=y_2$. A little bit of thought will tell you that $G$ affords two orbits on $X\times X$, one contains all the diaognal elements, and the other contains all the non-diagonal elements.
This means $|G\backslash X\times X| = 2$ and we get the desired result.
