I have two $\mathbb{C}$-linear semisimple tensor categories $C$ and $D$. Let $K(C)$ and $K(D)$ be their Grothendieck groups. I have a specific ring homomorphism $f \colon K(C) \to K(D)$ that I would like to realize as coming from a monoidal functor $F \colon C \to D$. What I know is that for every simple object $X \in C$, $f([X])$ is the genuine class of an object in $D$ (not just a virtual object). 


The tensor products on both $C$ and $D$ have a symmetry, so $K(C)$ and $K(D)$ are commutative rings. I do know that there is no monoidal functor that preserves the symmetry, but I don't care about preserving the symmetry.


Can this always be done or is there some kind of obstruction I need to consider?


Edit: Since there are many adjectives you can place in front of monoidal functor, I would like the version where $F(X \otimes Y) \cong F(X) \otimes F(Y)$ (satisfying the appropriate axioms).


Edit 2: In light of the Simon Henry's comment, let me add that both categories are linear over the complex numbers and that the endomorphisms of a simple object are just scalar multiples of the identity.



Lifting ring homomorphism of Grothendieck rings to functor of semisimple categories
Without the symmetry it’s easy to produce oodles of examples where a map of fusion rings doesn’t lift to the category level. The simplest example is probably $\mathrm{Vec}(G,w)$ the category of G-graded vector spaces with associator given by a 3-cocycle $w$. The fusion ring has an automorphism for any automorphism of G, but this only lifts to the category when it preserves $w$ in cohomology.


Giving a symmetric example is harder, but I think the following works. Let $D\_8$ be the $8$-element dihedral group, and let $\mathrm{Rep}(D\_8)$ be its category of representations. The fusion ring has an $S\_3$ of automorphisms which permute the three non-trivial invertible objects. (Side note: $\mathrm{Rep}(Q\_8)$ has the same fusion ring, and there this $S\_3$ comes from an action on $Q\_8$ and so lifts to the category level.) However, only one of the non-trivial elements of this $S\_3$ lifts to the category level. Morally the reason is simple, the 1-dimensional representations each have a kernel, two of them have kernel isomorphic to the Klein 4-group while one of them has kernel isomorphic to $\mathbb{Z}/4\mathbb{Z}$ and the only the permutation swapping the two Klein 4-group ones will lift to the category level. Unfortunately this isn't quite a proof because it's not clear how to describe the kernel of the representation using only the tensor category structure (and not the symmetric tensor category structure).


Nonetheless you can see that the other permutations don't lift by following the classification of Tambara-Yamagami categories in Section 9.2 of [Etingof-Nikshych-Ostrik](https://arxiv.org/abs/0909.3140), but it's a little technical. Namely, each 1-dimensional rep generates a $\mathrm{Rep}(\mathbb{Z}/2\mathbb{Z})$ category, and the 2-dimensional rep is fixed by tensoring on either side by this category, so this exhibits $\mathrm{Vec}$ as a bimodule category over $\mathrm{Rep}(\mathbb{Z}/2\mathbb{Z})$, and for the Klein 4-group kernel reps this bimodule is invertible while for the $\mathbb{Z}/4\mathbb{Z}$ kernel reps this bimodule is not invertible. Since this was all stated in the language of tensor categories, there's no tensor autoequivalence which mixes the two kind of 1-dimensional objects.


