I am trying to understand how to get all non-equivalent permutation representations of some transitive subgroup $T$ in $S_n$. The example in $S_4$ (see below) made me realize, that I have misunderstood some concepts, but cannot find where my mistake is. Any help or links would be appreciated.
Let $D_4$ (dihedral group of ord 8). It has three non-conjugate classes of subgroups isomorphic to $\mathbb{Z_2}$, two of which, say $C_1$ and $C_2$, are core-free. Thus, by letting $G$ act on the coset spaces $D_4/C_1$ and $D_4/C_2$, respectively, we get two faithful permutation representations of $D_4$ on $4$ objects, say $\phi_1$ and $\phi_2$. Thus, we can treat $\phi_1$ and $\phi_2$ as subgroups of $S_4$, say $H_1$ and $H_2$ (clearly both isomorphic to $D_4$).
Since $C_1$ and $C_2$ are not conjugate in $D_4$, it follows that $\phi_1$ and $\phi_2$ are not permutationally equivalent. (see Transitive G-sets $G/A$,$G/B$ Isomorphic iff $\exists x$ st. $x^{-1}Ax = B$). I thought this implies that $H_1$ and $H_2$ are two non-conjugate subgroups of $S_4$, both isomorphic to $D_4$. However, in $S_4$ there is only one conjugacy class of subgroups isomorphic to $D_4$, implying that all subgroups of $S_4$ which are isomorphic to $D_4$ are permutationally equivalent.
I have read the notions of permutation equivalence in Dixon and Mortimer, "Permutation groups" (pages 21-22) as well as in a related post here Example in which a normal subgroup acts non-equivalent on its orbits
But it makes it clear (by referring to Exercise 1.6.17) in Dixon and Mortimer's book that permutation equivalence is not the same as permutation isomorphism. It can be proved that two subgroups of $S_n$ are conjugate in $S_n$ if and only if they are permutation isomorphic.
The images of the two inequivalent representations $\phi_1$ and $\phi_2$ of $D_4$ that you mentioned are permutation isomorphic. This is because they are there is an automorphism $\alpha$ of $D_4$ such that $\phi_1(g) = \phi_2(\alpha(g))$ for all $g \in G$.