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**.