Question: Let $G$ be a finite group. Is it true that there is a subgroup $U$ inside some symmetric group $S_n$, such that $N(U)/U$ is isomorphic to $G$? Here $N(U)$ is the normalizer of $U$ in $S_n$.
Background: If true, this would for instance give a trivial proof of the Fried-Kollar Theorem that every finite group is the full automorphism group of a number field.
Results: If $U\le S_n$ acts regularly with respect to the natural action of $S_n$, then $N(U)/U\cong\text{Aut}(U)$. However, many finite groups are not the automorphism group of another finite group, like most cyclic groups. On the other hand, it is easy to get $N(U)/U\cong G$ for each abelian $G$ by choosing $U$ a direct product of semidirect products $C_{p_i}\rtimes C_{m_i}$ for suitable distinct primes $p_i$ and divisors $m_i$ of $p_i-1$, with the natural intransitive action of $U$ with orbit lengths $p_1, p_2,\dots$.
