In studying triangulated categories, some authors require the shift functor $T: \mathcal D \rightarrow \mathcal D$ to be an autoequivalence, whereas others require it to be an automorphism (i.e. strictly invertible). Unfortunately, I couldn't find any reference which clarifies whether the two requirements are actually equivalent or not. Indeed, I think they are, so, abstracting a little, here is my question: given an autoequivalence $T: \mathcal C \rightarrow \mathcal C$ of an arbitrary category $\mathcal C$, is it possible to find a functor $T': \mathcal C \rightarrow \mathcal C$ which is isomorphic to $T$ and is also an automorphism of $\mathcal C$?


More generally, one could ask if a similar result is true for functors between categories whose object sets are of the same cardinality...


Thanks in advance!



Is it possible to make an autoequivalence of categories an automorphism?
In practice one may always assume that such a shift is an automorphism instead of an auto-equivalence. But for that one also has to modify the category with an equivalence (but this is OK for applications):


Let $F : C \to C$ be an equivalence of categories. Define the following category $C'$: Objects are sequences of objects $(X\_n)\_{n \in \mathbb{Z}}$ with $X\_n \in C$, together with isomorphisms $\alpha\_n : F(X\_n) \cong X\_{n+1}$. It is clear how to define the morphisms. There is a canonical *automorphism* $F' : C' \to C'$, which is just a shift $(X\_n,\alpha\_n)\_n \mapsto (X\_{n+1},\alpha\_{n+1})\_n$. Besides the evaluation at $0$ yields a functor $C' \to C$. It is easy to check that this is an *equivalence* of categories which makes the diagram


$$\begin{matrix} C' & \stackrel{F'}{\rightarrow} & C' \\\\ \downarrow && \downarrow \\\\ C & \stackrel{F}{\rightarrow} & C \end{matrix}$$


commute up to natural isomorphism of functors.


To put this into more global terms: Let $\mathcal{A}$ be the $2$-category of all categories equipped with an auto-equivalence, and $\mathcal{A}'$ the full $2$-subcategory of $\mathcal{A}$ consisting of those categories equipped with an automorphism. Then $\mathcal{A}' \hookrightarrow \mathcal{A}$ is an $2$-equivalence of $2$-categories.


