Automorphic and geometric equivalence for commutative associative linear algebras

Let $\Theta$ be the variety of all commutative associative linear algebras over a fixed field $k$ of characteristic $0$. The group $\mathfrak{A/Y}$ is isomorphic to $\mathrm{Aut}k$.

Automorphic–geometric equivalence conjecture. Despite the fact that

$$\mathfrak{A/Y}\cong \mathrm{Aut}k,$$

automorphic equivalence coincides with geometric equivalence in this variety.

This claim concerns a case where the automorphism group $\mathfrak{A/Y}$ is nontrivial, yet the two equivalence notions are expected to agree. The source does not provide evidence of resolution.