Trying to think a bit about an MO question\*, and not being experienced in category theory, I happened to ask myself the following question that I'm quite sure would be pretty elementary for the non-layman in the field.
Of course feel free to close in case it's too localized or anyway not suited for MO.



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> Given categories $\mathcal{A}$ and $\mathcal{B}$ and a functor $F:\mathcal{A}\to\mathcal{B}$, is it the case that every functor naturally equivalent to $F$ is of the form $K \circ F \circ H$ for $H$ an autoequivalence of $\mathcal{A}$ and $K$ an autoequivalence of $\mathcal{B}$ ?
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\*) [the [one](https://mathoverflow.net/questions/56887/rigidity-of-the-category-of-schemes) on the rigidity of the category of schemes, but it's not relevant here]



Equivalent functors (elementary question)
Let $\mathcal{A}$ have objects $0$ and $1$, with morphisms $u\_i:i\to i$ for $i\in\{0,1\}$ satisfying $u\_i^2=1\_i$, and no other non-identity morphisms. Let $\mathcal{B}$ be a group $G$, regarded as a category with one object. Then for each pair $g=(g\_0,g\_1)\in G^{2}$ with $g\_0^2=g\_1^2=1$, we have a functor $F\_g:\mathcal{A}\to\mathcal{B}$. Moreover, $F\_g$ is isomorphic to $F\_h$ iff ($g\_0$ is conjugate to $h\_0$ and $g\_1$ is conjugate to $h\_1$). On the other hand, $F\_g$ is related to $F\_h$ by autoequivalences if and only if there is an auomorphism $\alpha$ of $G$ with (($h\_0=\alpha(g\_0)$ and $h\_1=\alpha(g\_1)$) or ($h\_0=\alpha(g\_1)$ and $h\_1=\alpha(g\_0)$)).


Now suppose we have elements $u,v\in G$ where $v$ is conjugate (but not equal) to $u$. We find that $F\_{(u,u)}$ is isomorphic to $F\_{(u,v)}$, but is not related to it by autoequivalences. 


