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Isn't it the case that a suitable property is "$\otimes$ has a right adjoint", per your answer to Monoidal categories whose tensor has a left adjoint? Or do you explicitly want a property in the form "every object is a monoid [...]"?
@QiaochuYuan: thank you! This is a very good answer (along with the argument via Day convolution that was here previously). It's tricky to know which answer to accept, because the process of developing the proof was very much a collaboration: I wish I could accept both answers. I've decided to accept Tim's answer, simply because it contains the full proof that $\mathscr V$ has finite products, but I appreciate that this answer already contains many of the key ideas. Sorry, and thank you again!
Thanks, this is a very elegant argument. To check my understanding: if we require that the tensor and unit have left adjoints, then the unit will be terminal, and the argument demonstrating that $X \otimes Y \cong X \times Y$ carries through as written, establishing that finite products do exist in $\mathscr V$, without having to take it as an assumption, right? In this case, cartesian categories may be characterised as monoidal categories such that the tensor and unit are right adjoint. That the left adjoints are the diagonals follows automatically.
@NoahSnyder: the concept is intended to be a weakening of the notion of cartesian category (where the cartesian product is right adjoint to the diagonal). However, one could just as well ask for a right adjoint instead, which would correspond to the cocartesian setting. I'm happy to know of references for either.
Higher operads and multicategories are mentioned in Leinster's Higher Operads, Higher Categories, though the focus there is on defining higher categories using the machinery of operads.
From the paper: "(Our nomenclature draws on one of the basic examples of a Galois object: if $k \subset K$ is a Galois field extension, and $\mathscr A$ is the category of intermediate field extensions, then $K$ is Galois in $\mathscr A^{\text{op}}$.)"
@NoahSchweber: the morphisms of a Lawvere theory are entirely abstract. Lawvere is taking the morphisms $X \to 2$ to be predicates in a first-order logic, but their semantic intent is entirely opaque to the cartesian category. You could take the morphisms to be whatever you wanted.
@NoahSchweber: could you elaborate? It seems to me that the category $\mathbf C$ Lawvere is defining in the paper mentioned is a 2-sorted Lawvere theory (with the sorts being $A$ and $2$).