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I struggle to see how this is research level - the first hit for "automorphisms of finite commutative groups" on google scholar gives me the American Mathematical Monthly article in the answer below. This article states that the result is from at least 1907.
It may be too restrictive a case to be useful, but any group of Euler characteristic $-1$, e.g. a 3-generator 1-relator group, cannot be embedded with finite index as a proper subgroup of any torsion-free group.
@YCor Ah, right. This has no algebraic consequences, as far as I know. I've just always found it an interesting quirk :-) The original motivation was indeed the conjugacy problem, but this is of course superseded by hyperbolicity. (Moreover, the conjugacy problem follows from Lyndon and Schupp's Theorem 5.3, with Theorem 5.5 really just tying up a loose end.)
@YCor I'm not sure what you mean. I've updated the statement to make it more precise, in the language of Lyndon and Schupp. The statement is false for $C'(1.6)$ presentations, as the 2-layer case is possible here (I can add an example if it would be helpful).
Twisted conjugacy generalises to "double-twisted conjugacy", which relates to the Generalised (or Non-Homogeneous) Post Correspondence Problem. This was originally posed for free monoids, which is why I mention it, but has been studied for groups (arxiv.org/abs/1310.5246, arxiv.org/abs/2211.12158). For groups, this takes two group homomorphisms $f, g:H\to K$ and a pair of elements $u, v\in K$, and asks if there exists a non-trivial word $x\in H$ such that $uf(x)v=g(x)$. The case of $u=v^{-1}$ corresponds to the situation here, and is undecidable for free groups.
Are you aware of Osin's construction of groups with prescribed conjugacy classes? It is on the arXiv here. This can be applied to give finitely generated simple groups which contain each of the subgroups you wish, and such that all elements of the same order are conjugate. (So misses on a few counts, but suggests hope...) I suspect promoting the finite generation to finite presentability in his construction would be extremely difficult though.
Going up a dimension from 2 to 3, I believe it is open whether 3-manifold groups satisfy Kaplansky's unit conjecture. Indeed, it has only recently been proven that they satisfy Kaplansky's zero-divisor conjecture (which is weaker) - this is due to Linton and Kielak (arxiv).