One suggested variant was to ask whether there are any interesting adjunctions between the category $FinGpd$ of finite groupoids and the category $FinHilb$ of finite-dimensional Hilbert spaces, which is still undefined. The answer is still no.
First let me suppose that the morphisms of $FinHilb$ are all linear maps. Then the automorphism group of $\mathbb C \in FinHilb$ is $\mathbb C^\ast$, a divisible abelian group. As such, it admits no nontrivial homomorphisms to any finite group. So if $F: FinHilb \to FinGpd$ is a functor, then the induced map $Aut(\mathbb C) \to Aut(F\mathbb C)$ is trivial; in particular, $F(1) = F(2)$ where $1,2: \mathbb C \to \mathbb C$ are the scalar maps. Moreover, if $F$ is a right adjoint, then it preserves abelian group objects, so we may subtract $1$ from $2$ to see that $F(0) = F(1)$. From this it follows that $F$ is constant at the terminal object. A similar argument using cogroup objects shows there are no interesting left adjoints $FinHilb \to FinGpd$.
On the other hand, if the morphisms of $FinHilb$ are chosen differently, then it's unlikely that $FinHilb$ has finite products / finite coproducts, so there can't be a nontrivial right / left adjoint functor $FinGpd \to FinHilb$.
The only exception I can think of is if one takes the morphisms of $FinHilb$ to be contractive linear maps, i.e. those maps $f$ such that $\|f(x)\| \leq \|x\|$. With this definition, $FinHilb$ still doesn't have finite products, so there can't be a right adjoint functor $FinGpd \to FinHilb$, but it has finite coproducts. But any right adjoint functor $FinHilb \to FinGpd$ will still identify all nonzero scalars for the same reasons as above, which I think already makes any such functor uninteresting.