Burnside's Lemma / Counting Formula says that the number of orbits of an action is equal to the average number of fixed points of the acting permutations. In my case, I'm particularly interested in the sizes of the orbits themselves for a particular action.

Is there a result as general as Burnside's Lemma but that deals with the sizes of orbits? Or, more specifically, does the following setup seem familiar to anyone?

**Additional Motivation:** The motivation for my question is something like the following.

Suppose we have a set of $m$ sharply transitive permutations $\Pi = \{\pi_1, \ldots, \pi_m\}$ on a set $X$. Here I mean that for $x,y \in X$, there is a unique $\pi \in \Pi$ such that $\pi(x) = y$.

The permutations cooperate "nicely" in the following sense. Let $A$ be an $m \times m$ matrix with off-diagonal entries drawn from $\{1, \ldots, m\}$ such that for $i \neq j$ whenever $\pi_i (x) = \pi_j (y)$, it follows that $\pi_j (x) = \pi_{A(i,j)} (y)$.

Let $G$ be the permutation group generated by $\Pi$. In the general case I'm working on, I really only know $A$, and I'm shooting for a statement of the form: "If $A$ has a certain property, then $|G(x)|$ is divisible by $m$."

The origins of the matrix $A$ while of course essential to proving anything like this statement remain sufficiently messy that I'd rather not get into it.