Group action on a Cartesian product Let $G$ be a finite group acting on a finite set $X$. Then naturally $G$ acts on $X \times X$ by $g.(x,y)=(g.x,g.y)$. Is there any way to find the number of orbits of the action of $G$ on $X\times X$ using the action $g$ on $X$? Are they related?

(Not a complete answer, but just an instance of the particular case mentioned in the comments.)
Let's call "$\star$" the induced action of $G$ on $X\times X$. Then, for $\bar x:=(x_1,x_2)\in X\times X$, the pointwise stabilizer reads:
\begin{alignat}{1}
\operatorname{Stab}_\star(\bar x) &= \{g\in G\mid g\star\bar x=\bar x\} \\
&= \{g\in G\mid (gx_1=x_1)\wedge(gx_2=x_2)\} \\
&=\operatorname{Stab}(x_1)\cap\operatorname{Stab}(x_2) \\
\tag 1
\end{alignat}
As an instance of the particular case mentioned by @TomGrubb in the comments, let's take $G=S_n$ and $X=\{1,\dots,n\}$, and as base action the natural one. Then, by $(1)$, $\operatorname{Stab}_\star((i,j))=\operatorname{Stab}(i)\cap\operatorname{Stab}(j)$, whence $\left|\operatorname{Stab}_\star((i,j))\right|=(n-1)!$ for $j=i$, and $\left|\operatorname{Stab}_\star((i,j))\right|=(n-2)!$ for $j\ne i$. Therefore, there are precisely two orbits for the action "$\star$"$^\dagger$, i.e. twice the number of orbits (one) of the (transitive) base one.

$^\dagger$By the orbit-stabilizer theorem, $\left|\operatorname{Stab}_\star((i,j))\right|=(n-1)!$ implies $\left|O_\star((i,j))\right|=n$, and $\left|\operatorname{Stab}_\star((i,j))\right|=(n-2)!$ implies $\left|O_\star((i,j))\right|=n(n-1)$. But the set of orbits is a partition of the acted on set, whose size is $n^2$, whence $kn+ln(n-1)=n^2$ or, equivalently, $k+l(n-1)=n$. For $n=2$, this yields $k+l=2$; for $n>2$, $l=\frac{n-k}{n-1}$ integer implies $k=1$, which in turn implies $l=1$, and then again $k+l=2$. So, the action "$\star$" has two orbits for every $n\ge 2$.
