# Is $\text{Aut}_{\text{br}}(C)$ braided?

Let $$C$$ be a braided monoidal category. The category $$\text{Aut}_{\text{br}}(C)$$ of braided monoidal autoequivalences of $$C$$ is monoidal with tensor product functor given by the composition $$\circ$$.

Is it a braided category as well?

If $$G$$ is an abelian group, let $$C$$ be the category whose objects are the elements of $$G$$ and with only identity morphisms ($$C$$ is a discrete category). $$C$$ is symmetric monoidal via the product in $$G$$. The category $$\text{Aut}_\text{br} (C)$$ is again discrete, it's objects are the automorphisms of $$G$$ ($$\text{Aut}_\text{br} (C)$$ is constructed out of $$\text{Aut}(G)$$ in the same way as $$C$$ out of $$G$$). Being descrete, it can be braided monoidal only if $$\text{Aut}(G)$$ is commutative, but for a general abelian $$G$$ it is not true. So the answer to your question in no.