Groups with no proper non-trivial fully invariant subgroup

Question

Let $G$ be a finite group. A subgroup $H$ of $G$ is said to be characteristic if $\phi(H)\subseteq H$, $\forall \phi \in \operatorname{Aut}(G)$ and fully invariant if $\phi(H)\subseteq H$, $\forall \phi \in \operatorname{End}(G)$.

A group $G$ is said to be characteristically simple if the only characteristic subgroups are $G$ and the trivial subgroup. It is possible to characterize characteristically simple groups: a group is characteristically simple if and only if it is the direct product of isomorphic simple groups (see chapter 2, section 1, of the book Finite Groups by Daniel Gorenstein).

Following J.D. Reid, "On rings on groups", Pacific J. Math., Vol. 53, No. 1, 1974, I call strongly irreducible a group $G$ if the only fully invariant subgroups are $G$ and the trivial subgroup.

My question: is it possible to characterize strongly irreducible finite groups?

Answer 1

Corollary 53.57 in Hanna Neumann's Varieties of Groups proves that a finite verbally simple group must be characteristically simple, hence of the form $S^n$ for some simple group $S$. (Also proven in Chain Markov's answer to Does there exist some sort of classification of finite verbally simple groups?)

If $G$ is finite and strongly irreducible, then it is verbally simple (because verbal implies fully invariant), and hence characteristically simple. Conversely, if $G$ is finite and characteristically simple, then because every fully invariant subgroup of a finite group must also be characteristic, then $G$ is strongly irreducible.

So the nontrivial finite strongly irreducible groups are precisely the characteristically simple groups, that is the groups of the form $S^n$ with $S$ finite and simple and $n\geq 1$.