I think that an example of non-equivalent permutation sets is given by
$G=(\mathbb Z/p\mathbb Z)^n$ for $n>2$ (and $p$ a prime). Then the automorphism
group is $\mathrm{GL}_n(\mathbb Z/p\mathbb Z)$, the conjugacy classes are in
natural bijection with $G$ and the set of irreducible representations are in
bijection with the dual group (or dual $\mathbb Z/p\mathbb Z$-vector space). In
both cases there are only two orbits, one of length $1$ (the identity element
and the trivial representation respectively). The stabilisers for elements in
the non-trivial orbits are not conjugate: Mapping to $\mathrm{PGL}_n(\mathbb
Z/p\mathbb Z)$ map these two kinds of stabilisers two non-conjugate parabolic
subgroups (stabilisers of lines resp. of hyperplanes).
I have come across a fairly recent result which pertains to this question. It is in this paper:
Guralnick, Robert M.; Maróti, Attila; Pyber, László, Normalizers of primitive permutation groups, Adv. Math. 310, 1017-1063 (2017). ZBL1414.20002.
An arXiv version is here. The paper examines the situation when $U$ is primitive. They show that, in all but a finite number of situations, $|N(U)/U|<n$. Indeed, they strengthen this bound if you add in a particular infinite family. The main result is this one:
Theorem: Let $U$ be a primitive subgroup of $S_n$, and let $N=N_{S_n}(U)$. Then $|N/U|< n$ unless $U$ is an affine primitive permutation group and the pair $(n, N/U)$ is one of:
$$(3^4,O^−_4(2),(5^4,Sp_4(2)),(3^8,O^−_6(2)),(3^8,SO^−_6(2)),(3^8,O^+_6(2)),(3^8,SO^+_6(2)),(5^8,Sp_6(2)),(3^{16},O^−_8(2)),(3^{16},SO^−_8(2)),(3^{16},O^+_8(2)), \textrm{ or }(3^{16},SO^+_8(2)).$$ Moreover if $N/U$ is not a section of $\Gamma L_1(q)$ when $n=q$ is a prime power, then $|N/U|< n^{1/2}\log n$ for $n≥2^{14000}$.
@YCor's comment on the original question suggests that the primitivity assumption is not too onerous. It would be interesting (but probably very hard) to try and understand what might happen when $U$ is transitive and imprimitive.
Best Answer
${\rm S}_6$ is not the automorphism group of a finite group. See H.K. Iyer, On solving the equation Aut(X) = G, Rocky Mountain J. Math. 9 (1979), no. 4, 653--670, available online here.
This paper proves that for any finite group $G$, there are finitely many finite groups $X$ with ${\rm Aut}(X) = G$, and it explicitly solves the equation for some specific values of $G$. In particular, Theorem 4.4 gives the complete solution for $G$ a symmetric group, and when $n = 6$ there are no such $X$.