Let $G$ be a permutation group on a finite set $\Omega$ with orbits $\Omega_1,\ldots,\Omega_k$.
By a *transversal* we mean a set $\lbrace\omega_1,\ldots,\omega_k\rbrace$ with $\omega_j\in\Omega_j$ for each $j$.

For a transversal $\tau=\lbrace\omega_1,\ldots,\omega_k\rbrace$ and subset $H\subseteq G$, let $$H(\tau) = \bigcup_{h\in H} \lbrace\omega_1^h,\ldots,\omega_k^h\rbrace. $$

Define $N=N(G)$ to be the smallest integer such that for some transversal $\tau$ and some $H\subseteq G$ we have $|H|=N$ and $H(\tau)=\Omega$.

**Question: What is $N(G)$?**

Let $n=|\Omega|$ and $m=\max_{i=1}^k |\Omega_i|$. It is easy to see that $N(G)\ge m$. By taking an arbitrary transversal and random subsets $H$, one can show non-constructively that $N(G)\le m\log n$.

This is an ideal version of a question that arose about symmetry breaking in finite optimization problems.