Non-equivalent primitive permutation representations of finite groups with the same set of derangements

Given a finite group G and a maximal subgroup T of G, define $\tilde{T} := \cup_{g \in G} T^{g}$. Suppose that there exist two maximal subgroups M and L of G such that $\tilde{M}= \tilde{L}$. The question is to determine if this implies that M and L are conjugated in G. In this work we first prove that the answer is affirmative when G is a soluble group or when G is a symmetric or alternating group and M, L are either intransitive or imprimitive. Then we prove that if G is the special linear group of degree 2 over a field of characteristic 2 then the answer is negative by showing a pair of non-conjugated maximal subgroups for which $\tilde{M}= \tilde{L}$ holds. Finally we give an answer for all the sporadic groups.