Is it possible to make an autoequivalence of categories an automorphism? 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!
 A: 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.
