TITLE: How are the left and the right group of a bitorsor related?
QUESTION [7 upvotes]: This question arose from my answer to To what extent does a torsor determine a group: it turns out that I do not know one thing about it.
Let $G$, $G'$ be groups in some nice enough category (you may assume a topos, if you feel like that). Can one find a nice intrinsic simplification of the condition "there exists a $G$-$G'$-bitorsor"?
It is clear that a necessary condition for the existence of a bitorsor is that $G$ and $G'$ are locally isomorphic, i. e. there is an object $B$ with global support such that the groups $B\times G\to B$ and $B\times G'\to B$ over $B$ are isomorphic over $B$. Is this also sufficient?
Can one do better? By this I mean not quantifying over objects but rather concocting some condition out of $G$ and $G'$ alone?

REPLY [2 votes]: [I don't have enough 'points' to comment; below is really just a comment.]
If you consider instead your 'nice enough' category to be locales, then having a bi-torsor between two open localic groups G and G' implies that their toposes of sheaves are equivalent. Quite different localic groups can define the same topos, so my instinct is that there is no easy or natural way to determine (Morita) equivalence by inspecting the groups other than to tautologically give the definition. (Refer to Remark C5.2.14(d) for a reference to a concrete example, for the groupoid case at least.)
I was intrigued by the question because you seemed to be interested in working on things to do with torsors in more general categorical contexts. Since I have done some work on torsors (effectively Hilsum-Skandalis maps) in a cartesian category, which I think is the most general possible context, I hope you don't mind my providing a link to that work: 
http://www.christophertownsend.org/Documents/HilsumSkandalisFrobenius.pdf
The general gist of the question seems to be: what can we say about how group(oids) are related/constructed given information about how their categories of equivariant sheaves are related; I find this to be an interesting avenue and only know partial answers.