We can divide the classification, at least initially, into torsion and nontorsion parts.

Nontorsion we classify immediately. They are the submodules of $\mathbb Q$, by Todd's argument.

Torsion is more subtle. We can divide a torsion module into $p$-power torsion groups for each prime $p$. Let $A$ be a $p^n$-torsion group, then $A[p^n]/A[p^{n-1}]$ is an $\mathbb F_p$-vectors space that injects naturally into $A[p^{n-1}]/A[p^{n-2}]$.  (Where $A[N]$ is the $N$-torsion part of $A$.) This gives a filtration of $\mathbb F_p$-vector spaces. We can choose a basis that agrees with this filtration, and by choosing lifts of the basis vectors we get an isomorphism between any two groups with the same filtration. In particular, we can write it as a direct sum of cyclic subgroups and the special module $\mathbb Q_p/\mathbb Z_p$.

The tensor of any two cyclic $p$-power group is nontrivial. The tensor of any cyclic group with $\mathbb Q_p/\mathbb Z_p$ is trivial. $\mathbb Q_p /\mathbb Z_p$ tensor itself is trivial.

So a torsion symtrivial group is a direct sum of, for each prime, up to $1$ $p$-power cyclic group and arbitrarily many copies of $\mathbb Q_p/\mathbb Z_p$.

Moreover, a symtrivial group must have a symtrivial torsion subgroup. Indeed, let $a \otimes b \neq b \otimes a$ in the torsion subgroup, but $a\otimes b = b \otimes a$ when the whole things is tensored. Then the proof that that inequality holds uses only finitely many elements of the group. Consider the subgroup generated by the whole torsion module and those finitely many elements. The nontorsion part of it is a finitely generated submodule of $\mathbb Q$, thus $\mathbb Z$, thus projective, so the torsion and nontorsion part split. This is a contradiction.

Thus, a symtrivial module consists of the choice of a torsion symtrivial module, a  torsion-free symtrivial module, and an appropriate extension between them. Next we determine which pairs of a symtrivial torsion and non-torsion group are possible.


The non-torsion part of $A$, localized at $p$, is either $\mathbb Z_p$ or $\mathbb Q_p$. If it is $\mathbb Z_p$ then the exact sequence splits and thus the $p$-torsion part must be trivial. If it is $\mathbb Q_p$, the $p$-torsion part can be any symtrivial $p$-torsion group. If you take the  direct sum of any pair of a symtrivial torsion group and a symtrivial non-torsion group that satisfy the condition that, at every prime, either the torsion is trivial or the non-torsion is divisible, you get a symtrivial group, so there is always at least one example.

Thus all that remains is to find out what exts are possible. I'm not sure how to go about that.

Everything so far works exactly the same over a general Dedekind domain.