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Consider a group $G$. We can get to a monoidal category $\cal G$ with objects being the group elements, tensor product given by group multiplication and just the identity morphisms as morphisms.

Further, let $\cal C$ be a monoidal category.

Assume we have an injective group homomorphism $$G\to\text{Aut}(\cal C)$$ into the group of monoidal autoequivalences of $\cal C$.

How do we get a monoidal functor ${\cal G}\to\text{Aut}(\cal C)$? Here the tensor product on $\text{Aut}(\cal C)$ is given by composition.

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  • $\begingroup$ $\text{Aut}(C)$ is a 2-group, not a group. (It can even be upgraded to a 3-group with some effort.) $\endgroup$ – Qiaochu Yuan Jul 28 '18 at 21:34
  • $\begingroup$ Ok, does this help to answer the question? I am not familiar with 2-groups. Is it a group if we assume $\cal C$ to be strict? $\endgroup$ – user497572 Jul 28 '18 at 21:39
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    $\begingroup$ I don't think you're using the monoidal structure on $\mathcal{C}$ if the tensor product in $\mathrm{Aut}(\mathcal{C})$ is just the composition. If that's the case, you already have a monoidal functor (you only need to define further on the arrows, and $\mathcal{G}$ only has identity arrows !). This is monoidal because you start from a morphism so $F(x\otimes y) = F(xy) F(x)F(y) = F(x)\otimes F(y) $ $\endgroup$ – Max Jul 28 '18 at 22:08
  • $\begingroup$ The monoidal structure of $\cal C$ is important as we want to consider the group of monoidal (!) autoequivalences of $\cal C$, not just the autoequivalences. For what do we need the injectivity of the group homomorphism? And is your functor fulfilling the hexagon diagram? $\endgroup$ – user497572 Jul 29 '18 at 10:03
  • $\begingroup$ You don't have an idea or is my question just a little stupid? :) $\endgroup$ – user497572 Jul 30 '18 at 20:24

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