Burnside's lemma says that if a finite group $G$ acts on a finite set $X$ then the number of orbits equals the average number of fixed points: $$|X/G| = \frac{1}{|G|} \sum_{g \in G} |X^g|.$$ I was wondering if there's anything that can be said in general, say in terms of measures: $$\# X/G = \frac{1}{\mu(G)} \int_G \#X^g d\mu?$$ Here are two examples to show that something like this could work under certain hypotheses (as yet unknown to me).
Example 1. Let $G$ be the group of invertible affine linear transformations $x \mapsto ax+b$ of $\mathbf{R}$ and let $X = \mathbf{R}$. Then the natural action of $G$ on $X$ is transitive (just translate), so $\#X/G = 1$. Now since $$ax + b = x \iff x = \frac{b}{1-a} \quad (a \ne 1)$$ we see that $\#X^g = 1$ generically, so the "average" number of fixed points is 1 also.
Example 2. Let $G = \langle -1 \rangle$ this time and keep $X = \mathbf{R}$. There are infinitely many orbits, namely $\{0\}$ and $\{x, -x\}$ for all $x \ne 0$. Here, $1$ fixes everything while $-1$ fixes only 0. Under the usual conventions surrounding arithmetic with $\infty$, the average number of fixed points is $$\frac{1}{2}\big(1 + \infty\big) = \infty$$ which is, again, the number of orbits.