Suppose one has two locally finite quasivarieties $\mathcal{V}$ and $\mathcal{W}$.

Further suppose that:

$\mathcal{V}$ is a variety.

The finite algebras $\mathcal{V}_f$ are dually equivalent to $\mathcal{W}_f$.

Does it follow that $\mathcal{W}$ is also a variety?

**Background**: This question fits into the area of `Natural duality'. I am hoping someone can provide a counterexample, or perhaps provide additional conditions.

Here is a way to manufacture such examples. Let $\mathcal{V}$ be a locally finite variety such that (i) $\mathcal{V}$-epis are surjective, (ii) $\mathcal{V}_f$ has an injective cogenerator $A$. Define the algebra $B$ to have the same carrier as $A$, its finitary operations between all $\mathcal{V}$-morphisms $A^n \to A$. Further define $\mathcal{W} := \mathcal{Q}(B) = \mathbb{S}\mathbb{P}(B)$ to be the quasivariety generated by $B$. Then $\mathcal{V}_f$ is dually equivalent to $\mathcal{W}_f$ via the contravariant hom-functors for $A$ and $B$, endowed with additional algebraic structure.

e.g. Let $\mathcal{V} = \mathsf{Vect}(\mathbb{F})$ for a finite field $\mathbb{F}$ and take $A = \mathbb{F}$. Then $\mathcal{W}$ is essentially $\mathcal{V}$ again, where every linear combination of variables is viewed as an operation.