Finite probability spaces.

Interpretation: we can consider $A$ and $B$ as events on the probability space $\Omega$ (randomly pick a point $x\in\Omega$; the events are whether $x\in A$ and whether $x\in B$). We call $A$ and $B$ ``independent events'' if $P(A\cap B)=P(A)P(B)$, ie., if $P(x\in A\cap B)=P(x\in A)P(x\in B)$, or yet in other words, if $\vert A\cap B\vert/n=(\vert A\vert/n)(\vert B\vert/n).$ The intuitive meaning of the formula stated in the exercise becomes more evident in this context if we divide each side by $n$:

$$E(\frac{\vert A\cap B^\sigma\vert}{n})=\frac{\vert A\vert}{n}\cdot\frac{\vert B\vert}{n}.$$

This means $A$ and $B^{\sigma}$ behave like independent events in the expectation.

Enhanced hint. Show that if $x$ is randomly chosen from $\Omega$ and $\sigma$ is randomly chosen from a transitive group $G$ then the events $x\in A$ and $x\in B^{\sigma}$ are independent. (This is immediate if $G=S_n$ (why?) but remains true for any transitive group $G$.) Then use indicator variables to show how this translates into the expected intersection size of $A$ and $B^{\sigma}$.