Suppose $k$ is an algebraically closed field of characteristic zero and $A$ is a finitely generated commutative associative reduced $k$-algebra.

Suppose the group $\mathbb{Z}_2$ acts on $A$ in such a way that the induced action on maxSpec $A$ is free. (Side note: This is equivalent to the induced grading $A = A_0 \oplus A_1$ being a strong grading, i.e., $A_1^2 = A_0$.)

Let $S^2 A = (A \otimes A)^{S_2}$, where $S_2$ acts by swapping the factors in the tensor product. The action of $\mathbb{Z}_2$ on $A$ induces a diagonal action of $\mathbb{Z}_2$ on $A \otimes A$. Since this action commutes with the $S_2$ action, we have an induced action of $\mathbb{Z}_2$ on $S^2 A$.

Now let $(S^2 A)_{\mathbb{Z}_2}$ be the corresponding algebra of coinvariants. By definition, this is the quotient of $S^2 A$ by the ideal generated by elements of the form $(b - g \cdot b)$ for $g \in \mathbb{Z}_2$ and $b \in S^2A$.

My question is: Is $(S^2 A)_{\mathbb{Z}_2}$ isomorphic to $A^{\mathbb{Z}_2}$ (as an algebra)?

My reason for thinking this is true is the following intuition:

$S^2 A$ is the coordinate ring of the set of unordered pairs of points of $A$.

Thus $(S^2 A)_{\mathbb{Z}_2}$ is the coordinate ring of set of unordered pairs of points of $A$ that are invariant under the $\mathbb{Z}_2$-action.

Since the action of $\mathbb{Z}_2$ is free, unordered pairs of points of $A$ that are invariant under the $\mathbb{Z}_2$ action are just $\mathbb{Z}_2$-orbits.

Thus $(S^2 A)_{\mathbb{Z}_2}$ should be isomorphic to $A^{\mathbb{Z}_2}$, which is precisely the coordinate ring of the variety of $\mathbb{Z}_2$-orbits.

I think the above reasoning is rigorous on the level of maximal ideals. But I'd like to know that the rings are actually isomorphic.