Equivalent functors (elementary question) 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.

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 $\mathcal{B}$ ?

*)  [the one on the rigidity of the category of schemes, but it's not relevant here]
 A: 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.  
A: For:  "a functor $G$ naturally equivalent to $F$" I  mean "exist a isomorphism $\alpha: F \cong G$" and for "autoequivalence" I mean "isomorphism". If $F$ is injective on the objects you can define a functor $H: \mathcal{B}\to \mathcal{B}$ as follow:
On objects let $H(F(X)):=G(X)$ and $H(B)=B$ if $B$ isn't in the image of $|F|_0$ 
On morphisms let $H(f)=\alpha_Y\circ f\circ \alpha_X^{-1}$ for $f: F(X)\to F(Y)$ 
let $H(f)= f\circ \alpha_X^{-1}$ for $f: F(X)\to B'$ where $B'$ isn't in the image of $|F|_0$.
let $H(f)=\alpha_Y\circ f$ for $f: B\to F(Y)$ where $B$ isn't in the image of $|F|_0$.
and $H(f)=f$,  if $f: B\to B'$  where $B$ and $B'$ aren't in the image of $|F|_0$.
Then $H$ is a functor and you have that $G=H\circ F$. 
PS: I dont like use choice-axiom on category-theory arguments 
