Does an equivalence of fusion categories depend on choice of simple objects within isomorphism classes? Let $C$ be a fusion category with simple objects $X_1,...,X_n $, and let $Y_1,...,Y_n$ be objects with each $Y_i$ isomorphic to $X_i$. Is there a monoidal auto-equivalence $F:C \rightarrow C $ which takes each $X_i$ to $Y_i$, and such that $F$ is naturally isomorphic (as a monoidal functor) to the trivial auto-equivalence?
It seems like this should be true since the structure of the fusion category is determined by $6j$-symbols and one should be able to get the same $6j$-symbols for every choice of representatives of simple objects by appropriately choosing representative morphisms between their tensor products. But how can one write down such a functor? 
 A: Let's forget about the monoidal structure for a moment.  $C$ is semisimple, and semisimplicity means that to give a functor out of $C$ is the same thing as specifying where it sends simple objects.  So that means that saying $\mathcal{F}(X_i)=Y_i$ determines a unique functor $\mathcal{F}$.  Now, what does it mean for this functor to be naturally isomorphic to the identity functor?  Well, again by semisimplicity, to give a natural transformation between functors is just to give its components on simple objects.  So fixing isomorphisms $\eta_i: X_i \cong Y_i$ determines a unique natural isomorphism $\eta: \mathcal{F} \rightarrow \mathcal{Id}$.
Now let's bring the monoidal structure back into the picture.  We have a functor $\mathcal{F}$ and a natural transformation $\eta: \mathcal{F} \rightarrow \mathcal{Id}$.  But $\mathcal{Id}$ has a canonical structure as a monoidal functor!  So you can just use "transport de structure" to move the monoidal structure on $\mathcal{Id}$ across the natural isomorphism to determine a monoidal structure $\mathcal{F}$.  In other words, there's no additional content to the monoidal statement.  Once you know that $\mathcal{F}$ is naturally isomorphic via $\eta$ to the identity functor, then there's a unique way to make $\mathcal{F}$ into a monoidal functor such that $\eta$ becomes a monoidal natural transformation.  This is completely the same argument as saying if you have a set X and a group G and a bijection $f: X \rightarrow G$ then there's a unique way to make X into a group such that $f$ is an isomorphism of groups.
