Let $D$ be a monoidal category (without the structure of a symmetric monoidal category), with unit object $Id$, and let $L$ be an invertible object in $D$, so that $L$ is dualizable and the pairing between $L$ and its dual $L^{*}$ is an isomorphism. (The situation I have in mind is that $D$ is the category of endofunctors of some category $C$ with composition giving the monoidal functor, and then $L$ is an autoequivalence of $C$.)

I want to put a monoid structure on $\coprod_{i \geq 0} D(Id, L^{\otimes i})$ by tensoring together two morphisms $\alpha: Id \rightarrow L^{\otimes i}$ and $\beta: Id \rightarrow L^{\otimes j}$ and then using the unit constraint on the domain $Id \otimes Id$ to get a morphism $\alpha \otimes \beta: Id \rightarrow L^{\otimes i +j }$.

Questions:

1. Does this construction make sense? I am a little worried that I have to fuss with the associativity constraints to really make sense of `$L^{\otimes i}$ '.

2. Presuming that the construction makes sense, is the resulting product commutative? To try to check this, I would (and did) attempt the following: Given a product of two morphisms $\alpha \otimes \beta: Id \rightarrow L^{\otimes i} \otimes L^{\otimes j}$, tensor it on the right with $L^{\otimes -j}$ and on the left with $L^{\otimes j}$. This produces a morphism $L^{\otimes j} \otimes Id \otimes Id \otimes L^{\otimes -j} \rightarrow L^{\otimes j} \otimes L^{\otimes i} \otimes L^{\otimes j} \otimes L^{\otimes -j}$. Now use the unit and pairing on the domain and the pairing on the codomain to identify this with a morphism $Id \rightarrow L^{\otimes j} \otimes L^{\otimes i}$. Is this morphism $\beta \otimes \alpha$? And could this show that $\alpha \otimes \beta= \beta \otimes \alpha$, once the appropriate identifications of domain and codomain have been used?