Do the sizes of the orbits determine a permutation representation? Suppose $\pi: G \to \mathrm{Sym}(X)$ and $\pi': G \to \mathrm{Sym}(X')$ are two permutation representations with, respectively, orbits $X_i (i \in I)$ and $X_j (j \in J)$. Suppose further that there is a bijection $\sigma: I \to J$ such that $|X_i| = |X_{\sigma(i)}|$. Does it follow that $\pi$ and $\pi'$ are equivalent, i.e. that there is a bijection $\beta: X \to X'$ such that $\beta(\pi(x)) = \pi'(\beta(x))$ for any $x \in X$?
I have a strong feeling that the answer is affirmative; clearly there is some bijection from $X$ to $X'$, since, by the hypothesis, they have the same cardinality. But I'm not totally sure about how to find the desirable $\beta$. Any clues? 

Let's think about what this means in the case that there is only one orbit in each of $X$ and $X'$ (the general case is easily reduced to that case anyways).  In that case, if you let $H$ be the stabilizer of a point of $X$ and $H'$ be the stabilizer of a point of $X'$, $\pi$ is equivalent to the action of $G$ on the coset space $G/H$ and $\pi'$ is equivalent to the action of $G$ on the coset space $G/H'$.  Two such coset spaces are equivalent representations of $G$ iff $H$ and $H'$ are conjugate as subgroups of $G$.
So, in this framework, your question becomes: if a group $G$ has subgroups $H$ and $H'$ such that $|G/H|=|G/H'|$, then must $H$ and $H'$ be conjugate?  Can you answer the question yourself now?  The answer is hidden below.

 The answer is no.  For the smallest counterexample, take $G=\mathbb{Z}/2\times\mathbb{Z}/2$ and let $H$ be generated by $(1,0)$ and $H'$ be generated by $(0,1)$.

