Monoidal categories in which $\mathrm{Aut}(X \otimes Y) \cong \mathrm{Aut}(X) \sqcup \mathrm{Aut}(Y).$

Question

I'm looking for examples of monoidal categories $\mathbf{C}$ such that one of the following two statements holds.

1. For all objects $X$ and $Y$ of $\mathbf{C},$ $\mathrm{Aut}(X \otimes Y) \cong \mathrm{Aut}(X) \sqcup \mathrm{Aut}(Y)$ in the category of groups.

2. For all objects $X$ and $Y$ of $\mathbf{C},$ $\mathrm{End}(X \otimes Y) \cong \mathrm{End}(X) \sqcup \mathrm{End}(Y)$ in the category of monoids.

Note that the coproduct in the category of groups (monoids) is different from the coproduct in the category of groupoids (categories), or more precisely that the corresponding forgetful functors do not preserve all coproducts. This kind of motivates my question; I'm trying to understand the various natural ways in which the coproduct of groups (monoids) arises.

Answer 1

This is almost never satisfied since the endomorphisms (automorphisms) of $X$ resp. $Y$ commute when extended to $X \otimes Y$. It is more reasonable to ask if $\mathrm{End}(X \otimes Y) = \mathrm{End}(X) \times \mathrm{End}(Y)$ holds. For example, this holds in the free monoidal category on a category. It fails in most symmetric resp. braided monoidal categories because we cannot express the symmetry $X \otimes X \cong X \otimes X$ as a tensor product of two endomorphisms of $X$.

For closed symmetric monoidal categories it makes more sense to ask if $\underline{\mathrm{End}}(X \otimes Y) = \underline{\mathrm{End}}(X) \otimes \underline{\mathrm{End}}(Y)$ holds. For example, this holds in the category of finite-dimensional vector spaces over a field (but not in the full category of vector spaces). More generally, let $C$ be a closed symmetric monoidal category and $X,Y,X',Y' \in C$. There is a canonical homomorphism $$\underline{\mathrm{Hom}}(X,X') \otimes \underline{\mathrm{Hom}}(Y,Y') \to \underline{\mathrm{Hom}}(X \otimes Y,X' \otimes Y').$$ It is an isomorphism when $X$ and $Y$ are dualizable, because then both sides identify with $X^* \otimes Y^* \otimes X' \otimes Y'$. In particular, for two dualizable objects $X,Y$ we have $\underline{\mathrm{End}}(X \otimes Y) = \underline{\mathrm{End}}(X) \otimes \underline{\mathrm{End}}(Y)$.