Conjugation of regular subgroups in the holomorph

Question

Given a finite group $G$, denote by $Hol(G)$ its holomorph and by $S(G)$ its group of permutations. It is well known that for regular subgroups $N,M\le Hol(G)$, being isomorphic or being conjugate are two equivalent data. Assume that this holds, i.e. let $M,N\le Hol(G)$ regular such that $\sigma^{-1}N\sigma=M$ for some $\sigma\in S(G)$. Being $N,M$ in the holomorph, is it restrictive to assume that such $\sigma\in Aut(G)$?

Answer 1

I'm still not entirely sure about the formulation of the question, but the following is probably a counterexample: The holomorph of $C_9$ can be generated by the normal subgroup $N=\langle (1,2,3,4,5,6,7,8,9)\rangle\cong C_9$ and the group $U=\langle (2,3,5,9,8,6)(4,7)\rangle$ that acts as the full automorphism group of $N$. Also in the holomorph is the regular subgroup $M=\langle (1,2,6,4,5,9,7,8,3)\rangle$, also isomorphic to $C_9$. As isomorphic regular groups, $N$ and $M$ are conjugate in $S_9$ (namely by $\sigma=(3,6,9)$), but they are not conjugate in the holomorph, and $\sigma$ does not normalize $N$ (and thus cannot induce an automorphism of $N$).