I want some finite set of data to which I can canoically associate a "group up to inner automorphism", and which can be constructed canoically from a "group up to inner automorphism". I have a few answers which satisfy the first requirement, but not the second.

**Give a topological space $X$.**The fundamental group of $X$ is a "group up to inner automorphism": you can get a group by picking a basepoint $p$, but given two different basepoints $p$ and $q$, there is not a canonical isomorphism $\pi_1(X,p)\to\pi_1(X,q)$, but any two automorphisms coming from a path from $p$ to $q$ differ by an inner automorphism. This answer is unsatisfying because there are many topological spaces one can pick, even if we restrict them to be $K(\pi,1)$'s.**Give a groupoid $X$**(this generalizes the first example by taking the fundamental groupoid). This answer is unsatisfying because again there are many (say, finite) groupoids corresponding to a given group up to inner automorphism.**Give a tensor category which is isomorphic to the category of finite dimensional representations of a finite group over an algebraically closed field of characteristic zero.**This is almost a good answer. It looks at first that specifying such a category is a finite amount of combinatorial data: we specify the isomorphism classes of objects and specify how the tensor product of any two of them decomposes. However we also have to specify isomorphisms $(A\otimes B)\otimes C\to A\otimes(B\otimes C)$, and this means specifying some matrix which depends on the specific bases of the isotypic components of the tensor products that we chose when specifying how they decompose. So it is unsatisfying as well, although it is the best I have come up with. It is almost good enough since it is easy to say what an equivalence between two such finite categories is in terms of matrices.**Give an orbifold whose coarse space is a single point.**This answer is unsatisfying because I am hoping the answer to this question will be useful for giving a nice definition of an orbifold. Also, by unraveling the definition of an orbifold, this answer really just reduces to (1) or (2) and is thus also unsatisfying.

is the identity(2) given $G$ and the associated data, I should be able to construct a group $G'$ and an isomorphism $G'\to G$ which is well-defined up to inner automorphism. $\endgroup$ – John Pardon Feb 15 '13 at 1:14unbasedhomotopy classes ofunbasedmaps? The construction of $BG$ depends on choices, as you say, but all such choices lead to isomorphic objects. $\endgroup$ – Mark Grant Feb 15 '13 at 9:25