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? 



Does an equivalence of fusion categories depend on choice of simple objects within isomorphism classes?
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.


