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@JoséAlejandroSamper, I thought when I updated the post that I saw how to easily recover nonevasiveness, but now I no longer see it. I'll delete my earlier comment claiming that it holds.
@JoséAlejandroSamper, as far as examples, look at the graph theory literature, under the name independent domination. A nontrivial example can be constructed by attaching a large number of pendant edges to each vertex of a complete graph. See e.g. Independent domination in graphs: a survey and recent results by Goddard and Henning.
fixed a problem pointed out by @Hailong Dao; hopefully improved clarity of exposition. I remark that non-evasiveness still seems to hold (per comments).
In particular, the Stadje paper mentioned by @usul may be found on JStor: jstor.org/stable/1427566 . The desired expected value is given by Equation 2.15 in the case p=1. (The author also finds formulas for higher moments.)
In answer to @NickGill's question: the answer is "yes". Indeed, if G and H have the same lattice of subgroups, then they are already both (super)solvable or both not (super)solvable. There are effective ways to detect solvability from the subgroup lattice: Roland Schmidt gave a characterization in terms of chains of modular elements with a flavor entirely similar to that of the definition. My own favorite characterization is the topological/combinatorial characterization of John Shareshian.
As far as subgroup of order 5, these groups have 1 normal subgroup, 1 conjugacy class of 5 sgs, and 1 conjugacy class of 25 sgs. (Added this comment when I fixed a 5 that should have been p, after @YCor pointed out a problem.)
