Is the finite inverse semigroup isomorphism problem GI-complete? Is the finite inverse semigroup isomorphism problem GI-complete? Here the finite inverse semigroups are assumed to be given by their multiplication tables.
Yes, the finite inverse semigroup isomorphism problem is GI-complete! This is a corollary of


  Theorem: Lattice isomorphism is isomorphism complete


from section 7.2 Lattices and Posets in


  Booth, Kellogg S.; Colbourn, C. J. (1977), Problems polynomially equivalent to graph isomorphism, Technical Report CS-77-04, Computer Science Department, University of Waterloo.


because a (semi-)lattice is also an (idempotent commutative) inverse semigroup.

Proof of theorem from technical report:


  It suffices to represent a graph uniquely as a lattice. Given a graph $G$ with $n$ vertices and $m$ edges, we define a lattice with an element for each vertex, an element for each edge, and two additional elements $\text{'}0\text{'}$ and $\text{'}1\text{'}$. Element $\text{'}1\text{'}$ dominates all others (the supremum), and element $\text{'}0\text{'}$ is dominated by all other elements (the infimum). An edge dominates exactly those vertices which are its endpoints. The result is a lattice which uniquely represents $G$.




The idea for this answer came from a discussion with vzn about sufficiently focused questions. The motivation to spend time on graph isomorphism at all also came from vzn's repeated prodding. J.-E. Pin asked in the comment whether there are any specific reasons to consider inverse semigroups. The idea was to have a structure slightly generalizing groups, which is GI complete. I wanted to better understand the relation between group isomorphism and graph isomorphism, but I fear this answer doesn't provide any insight of this sort.