Is this a correct characterisation of natural equivalence? In the category of functors of type $\mathcal{C}\rightarrow \mathcal{D}$ an arrow $\psi: F \Rightarrow G$ is defined as a collections of $\left(\psi_X: FX \rightarrow GX| X:\mathcal{C} \right)$ of compenents in $\mathcal{D}$, such that $\psi_Y \circ FX = GY \circ \psi_X$ and in case all the components are invertible, $\psi$ is called a natural equivalence or natural isomorphism of $F$ and $G$. My question is whether or not:

There exists a natural equivalence of $F,G: \mathcal{C}\rightarrow \mathcal{D}$, iff $FX \simeq GY \mbox{ in } \mathcal{D} \mbox{ whenever } X \simeq Y \mbox{ in }\mathcal{C}$

in other words:
$$F\simeq G \mbox{ in } \mathcal{C}^{\mathcal{D}} \Leftrightarrow FX \simeq GY \mbox{ in } \mathcal{D} \mbox{ whenever } X \simeq Y \mbox{ in } \mathcal{C}$$
Thanks
 A: $FX\simeq GX$ for all objects $X$ doesn't imply that $F\simeq G$ as functors. 
For example, let $\mathcal{C}$ be the category with two objects $X$ and $Y$, and only one morphism other than the identity morphisms, say $\alpha:X\to Y$. Then there are two functors $F$ and $G$ from $\mathcal{C}$ to (say) the category of real vector spaces with $FX=GX=FY=GY=\mathbb{R}$, $F\alpha=\operatorname{id}_{\mathbb{R}}$ and $G\alpha=0$, which are not isomorphic as functors.  
A: No. For example, let $C$ be a one-object category corresponding to a group $G$ and let $D = \text{Vect}$. Then a functor $C \to D$ is a linear representation of $G$, natural isomorphism of functors is isomorphism of linear representations, and the equivalence relation you define is instead isomorphism of underlying vector spaces. 
A: The answer is no.
From left to right:
For all $X \simeq Y \mbox{ in } \mathcal{C}$
$$F\simeq G \mbox{ in } \mathcal{C}^{\mathcal{D}} \Rightarrow FX \simeq FY \simeq GY \mbox{ in } \mathcal{D} $$
From right to left however:
For all $X \mbox{ in } \mathcal{C}$
$$ X \simeq X \mbox{ in } \mathcal{C}  \Rightarrow FX \simeq GX \mbox{ in } \mathcal{D}$$
but this is not enough to imply that $ F\simeq G \mbox{ in } \mathcal{C}^{\mathcal{D}}$
as Jeremy Rickard shows above
