2 minor clarification

In the case of a finite group $G$, a lot can be said by looking at $Fix_G(PPX)$, as it includes information on primitivity of the action on $X$, etc. E.g. if $G$ is doubly transitive on $X$, one can list all the possible examples of $H$ equivalent to $G$ using the classification of finite simple groups.

More generally, for primitive groups, one can use O'Nan-Scott theorem to partition such groups into few relatively well-understood classes, and, hopefully, derive the list you are looking for. As a toy example, consider $G\cong S_5$ acting on the set $X$ of pairs of {1,...,5}. There are just two nontrivial invariant graphs on $X$, the Petersen graph, and its complement. $G$ is the automorphism group of the Petersen graph, thus $H$ must be a subgroup of $G$. It follows by inspection that the only $H\neq G$, $G$~$H$ is (EDIT: actually, it could be that this $H$ is distinguished from $G$ by other orbits on sets, this still needs to be checked!) the index 2 subgroup in $G$, isomorphic to $A_5$.

For imprimitive groups, probably there is a reduction to the primitive case. (And needless to say, intranisitive case reduces to the transitive.)

1

In the case of a finite group $G$, a lot can be said by looking at $Fix_G(PPX)$, as it includes information on primitivity of the action on $X$, etc. E.g. if $G$ is doubly transitive on $X$, one can list all the possible examples of $H$ equivalent to $G$ using the classification of finite simple groups.

More generally, for primitive groups, one can use O'Nan-Scott theorem to partition such groups into few relatively well-understood classes, and, hopefully, derive the list you are looking for. As a toy example, consider $G\cong S_5$ acting on the set $X$ of pairs of {1,...,5}. There are just two nontrivial invariant graphs on $X$, the Petersen graph, and its complement. $G$ is the automorphism group of the Petersen graph, thus $H$ must be a subgroup of $G$. It follows by inspection that the only $H\neq G$, $G$~$H$ is the index 2 subgroup in $G$, isomorphic to $A_5$.

For imprimitive groups, probably there is a reduction to the primitive case. (And needless to say, intranisitive case reduces to the transitive.)