This is a follow up to my question What is the precise relationship between groupoid language and noncommutative algebra language?. I will briefly review some definitions; for details, a good place to look is Christian Blohmann, Alan Weinstein. Group-like objects in Poisson geometry and algebra. 2007. arXiv:math/0701499v1. And actually, there are two versions of my question, one for groupoids and the other for categories. So that I can avoid all analysis, I will restrict my attention to finite things; if you know the answer in, say, topological spaces, or smooth manifolds, or..., then I'm also interested.

A **category** is a span of sets $C = \{ C_0 \overset l \leftarrow C_1 \overset r \rightarrow C_0\}$ which is an algebra object in the category of $C_0,C_0$ spans. I.e. there are maps of spans $i: \{C_0 = C_0 = C_0\} \to C$ and $m: C \underset{C_0}\times C \to C$ making the usual diagrams commute. A category is a **groupoid** if additionally there is an involution ${^{-1}} : \{ C_0 \overset l \leftarrow C_1 \overset r \rightarrow C_0\} \to \{ C_0 \overset r \leftarrow C_1 \overset l \rightarrow C_0\}$ satisfying some condition. A category $C$ is **finite** if both $C_0$ and $C_1$ are finite.

A finite-dimensional algebra $A$ (over a fixed field $\mathbb K$) is **sesqui** if it is equipped with a bimodule ${_A \Delta _{A\otimes A}}$ and an "associativity isomorphism" $$\varphi: {_A \Delta _{A\otimes A}} \underset{A\otimes A}\otimes \bigl( {_A A _A} \underset{\mathbb K}\otimes {_A \Delta _{A\otimes A}}\bigr) \overset\sim\to {_A \Delta _{A\otimes A}} \underset{A\otimes A}\otimes \bigl( {_A \Delta _{A\otimes A}} \underset{\mathbb K}\otimes {_A A _A} \bigr) $$
of $A, A^{\otimes 3}$ bimodules, which satisfies a pentagon. There should also be a "counit" bimodule $_A \epsilon _{\mathbb K}$, some triangle isomorphisms, and some more equations.
A sesquialgebra is **hopfish** if a hard-to-write-down condition is satisfied; see Xiang Tang, Alan Weinstein, Chenchang Zhu. Hopfish algebras. 2006. arXiv:math/0510421v2. Let ${_A {\Delta^{\rm flip}} _{A\otimes A}}$ denote the bimodule $\Delta$ with the two right $A$-actions flipped. A sesquialgebra is **symmetric** if it comes equipped with a bimodule isomorphism $\psi: {_A \Delta _{A\otimes A}} \overset\sim\to {_A {\Delta^{\rm flip}} _{A\otimes A}}$ so that $\varphi,\psi$ satisfy two hexagons. A sesquialgebra is **finite** if $A,\Delta, \dots$ are finite-dimensional over $\mathbb K$.

Let $C = \{ C_0 \overset l \leftarrow C_1 \overset r \rightarrow C_0\}$ be a finite category. Then it gives rise to a finite symmetric sesquialgebra as follows. The algebra $A$ is given by the vector space $\mathbb K C_1$ with the convolution product (given on the basis by $a\otimes b \mapsto ab$ if $(a,b)$ is a composable pair of morphisms, and $a\otimes b \mapsto 0$ otherwise). The bimodule $\Delta$ is given as the vector space with basis all pairs $(a,b) \in C_1 \times C_1$ with $l(a) = l(b)$. I will let you work out the rest: the actions, the associator $\varphi$ and symmetrizer $\psi$, etc. If $C$ is actually a groupoid, then $\mathbb K C_1$ is hopfish. This construction extends to a 2-functor, and so sends equivalences of categories to Morita equivalences of sesquialgebras.

Question:It is well known that a groupoid $C$ cannot be recovered from the algebra $\mathbb K C_1$; compare for example the group with two elements, thought of as a groupoid with one object, and the set with two elements, thought of as a groupoid with only identity morphisms. But the examples I know can be distinguished by remembering the hopfish structure.

- Can a finite category be recovered from its symmetric sesquialgebra?
- If not, can a finite groupoid be recovered from its symmetric hopfish structure?
- Can an equivalence class of finite categories be recovered from the Morita-equivalence class of finite symmetric sesquialgebras?
- If not, do we at least have the corresponding statement for groupoids/hopfish algebras?

In a footnote, Blohmann and Weinstein suggest that they do not know the answers to the above questions. But that was three years ago; perhaps there has been more recent work?