When can you cancel the identity functor in a natural isomorphism? Suppose endofunctors $F, G : \mathbb C \to \mathbb C$ and the identity functor $1$. If we have an isomorphism $F \times 1 \cong G \times 1$, then under what conditions can we get $F \cong G$?
In Set, this seems to be fine so long as $F\emptyset \cong G\emptyset$. For any non empty set $X$, we have $FX \times X \cong GX \times X \Longrightarrow FX \cong GX$. However in other categories this condition doesn't seem to be enough.
Does anyone know some more general conditions on either $\mathbb C$ or $F,G$ that would let us cancel the $1$ like this?
 A: If $\mathbf C =\mathbf {Set}$, I'd say that you can cancel the identity as long as $F,G$ preserve copowers. In fact, denoting the singleton by $*$, $$F(*)\times *\cong G(*)\times *$$ implies that $F(*)\cong G(*)$; since every set is a copower of $*$, any functor that preserves copowers is uniquely determined (modulo equivalence) by its image on $*$.
In your question, you probably meant to construct the isomorphisms $FX\cong GX$ restricting $FX\times X\cong GX\times X$ to $FX\times x\cong GX\times x$, choosing $x\in X$ (for any $X$ that is nonempty). However at that point the naturality square, for $f:X\to Y$, is: $\require{AMScd}$ $$\begin{CD}
FX\times x @>\sim>> GX\times x\\
 @VVFfV @VVGfV \\
FY\times y@>\sim>> GY\times y
\end{CD}$$ that if $f(x)\neq y$ is not necessarily commutative.
EDIT. Actually, an isomorphism $\alpha:FX\times X\cong GX\times X$ induces an isomorphism $FX\times x\cong GX\times x$ just if one assumes that this diagram commutes: $$\begin{CD}
FX\times X @>\alpha>> GX\times X\\
 @VVV @VVV \\
X@>\operatorname{id}>> X
\end{CD}$$
