I was given the exercise to show that a $\pi_{\leq 1}(S^1)$ (fundamental groupoid of $S^1$)-module category is equipped with a natural automorphism of identity functor.
If $C$ is such a category, I need to produce an isomorphism $A \to A$ for every $A \in C$. My idea is to consider the isomorphism, for any $e^{i\theta}$ an object of $\pi_{\leq 1}(S^1)$, $$ A \mapsto (e^{i\theta}\triangleright A) \mapsto e^{-i\theta}\triangleright(e^{i\theta}\triangleright A) \cong (e^{-i\theta}\otimes e^{i\theta})\triangleright A = e^0\triangleright A\cong A $$ by the axioms of a module category, and I would then check naturality as well. But because this works for any $\theta \in [0,2\pi)$, I instead have a family of natural automorphisms. Technically this fulfills the requirement of the task, but the wording makes it seem as if there should be some canonical natural automorphism, which it appears I have not found.
Q: i) Does what I wrote down describe a natural automorphism of identity functor? ii) If so, is it what the question intends for, or is there something else more canonical?
Let $1 \in S^1$ be the basepoint, and let $\gamma$ be a path that starts at $1$, wraps around the circle once, and ends back at $1$. Then, it seems that the natural automorphism you want is $$A \xleftarrow{\cong} 1 \triangleright A \xrightarrow{\gamma \triangleright \mathrm{id}} 1 \triangleright A \xrightarrow{\cong} A.$$ The inverse map is given by the same construction, but with the reversed path $\bar{\gamma}$ in place of $\gamma$. Furthermore, naturality follows from the naturality of $1 \triangleright (-)$ and the fact $$(\mathrm{id} \triangleright f) \circ (\gamma \triangleright \mathrm{id}) = \gamma \triangleright f = (\gamma \triangleright \mathrm{id}) \circ (\mathrm{id} \triangleright f)$$ for any morphism $f$.