Do the usual examples of "dimensive" Lawvere theories form an exhaustive list of such theories?

Question

Call a Lawvere theory $T$ dimensive iff, letting $F_T : \mathbf{Set} \rightarrow \mathbf{Mod}(T)$ denote the free functor, we have the following.

• Every finitely generated $T$-algebra is free.
• From $F_T(J) \cong F_T(I)$ we may deduce $J \cong I$, for all finite sets $I,J$.

Motivation. If $T$ is dimensive, then every finitely-generated $T$-algebra has a well-defined dimension.

Examples.

• The initial Lawvere theory (whose models are sets).
• For each field $K$, the Lawvere theory of $K$-modules.

Question. Is this an exhaustive list of dimensive Lawvere theories?

Answer 1

This is getting too long for a comment, but here are some more examples:

• Modules over a division ring $K$ work.
• Also, pointed sets work.
• Affine spaces over a field, or more generally a division ring, work.

These examples come from here where Lawvere theories with all algebras free are considered.

Some observations from over there still apply. Your Lawvere theory must have at most one constant. If you want all algebras to be free (not just finitely-generated ones), then affine spaces over a division ring and sets themselves are the only examples with no constants, and it seems likely that pointed sets and vector spaces over a division ring are the only examples with constants.

Update If you follow the link above, you'll see that Keith Kearnes and collaborators now have a paper out showing that these are the only Lawvere theories for which every finitely generated algebra is free, regardless of the second condition! They all satisfy the second condition, too.