Maybe I'm going to make a dumb mistake here, since I'm writing this down rather quickly. Tell me what you think.

I think I claim this can happen all the time.

What about taking G an answer to be a finite simple group, letting C=Rep(G), and letting D=C^* _{Vyour question is given in http://arxiv.org/pdf/0810.0032, where C^*V means take theorem 1.1.}

Subcategories of the dual category relative to V=category double D(G) are given by pairs of Vector spacesnormal subgroups K,N in G which centralize each other, regarded as a module category over C via together with the fiber functor. Now, D will be pointed, as it will be equal to representations data of the dual hopf algebra a bicharacter K\times N \to C[G], which is commutativeC^\times.A pointed fusion category is necessarily equal to Rep(H,w), where H is a finite abelian group

So in particular if G has no normal subgroups and w is a cocycle. HoweverH does, this is actually then you're going to be trivialfind that D(G) has no nontrivial subcategories, because you already realized D as Rep(C[G]^while D(H) . So that means D is just Rep[H] for H abelianwill (one can take, and K=the normal subgroup in particular not simple. By definition, these share an invertible bimodule categoryH, namely VecN={id}, so and the bicharacter K\to C^* to be trivial I think this answers your questionguess.

Noah,

Maybe I'm going to make a dumb mistake here, since I'm writing this down rather quickly. Tell me what you think. I think I claim this can happen all the time.

What about taking G to be a finite simple group, letting C=Rep(G), and letting D=C^* _V, where C^*_V means take the dual category relative to V=category of Vector spaces, regarded as a module category over C via the fiber functor. Now, D will be pointed, as it will be equal to representations of the dual hopf algebra to C[G], which is commutative. A pointed fusion category is necessarily equal to Rep(H,w), where H is a finite abelian group and w is a cocycle. However, this is actually going to be trivial, because you already realized D as Rep(C[G]^). So that means D is just Rep[H] for H abelian, and in particular not simple. By definition, these share an invertible bimodule category, namely Vec, so I think this answers your question.

-david