Is a natural transformation uniquely determined by a single morphism? Let $C$ and $D$ be categories, let $F$ and $G$ be functors from $C$ to $D$, and let $\gamma$ and $\delta$ be natural transformations from $F$ to $G$.  Then my question is, if $\gamma_a=\delta_a$ for some $a\in C$, then is it necessarily true that $\gamma=\delta$? In other words, is a natural transformation uniquely determined by a single morphism?
If it’s not true in general, is it at least true if all the morphisms in $D$ are isomorphisms?  I think it might be implied by the naturality condition.
 A: No. Let $C$ be a discrete category. Then a natural transformation/isomorphism is just a $C$-indexed family of morphisms/isomorphisms of $D$. The naturality condition becomes vacuous.
A: Note $E^*$ the process of adjoining an element $*$ with its identity arrow to a category $E$. If we have distinct natural transformations $\alpha, \beta : F \Rightarrow G$, from functors $F,G : A \to B$, then we get functors $F^*,G^* : A^* \to B^*$ by mapping $*_A \mapsto *_B$. Likewise, we have natural transformations $\alpha_*,\beta_* $ which have an additional component added, $\alpha_* = \beta_* = 1_*$. These now share a common component, but are still distinct because the original were. Moreover if we take $B$ to be a groupoid, so will be $B^*$. 
Now, for a more 'concrete' counterexample, you could do the former with functors $F = G = 1 : BK \to BK$ for some abelian group $K$, and natural transformations corresponding to distinct elements $k,k' \in K$.  
There are however affirmative results "in this direction", the most famous being the Yoneda Lemma: if $F : C \to \mathbf{Set}$ is a functor from a locally small category to $\mathbf{Set}$ and $x \in C$, then we have a bijection
$$
Nat(hom_C(x,-),F) \simeq Fx
$$
given by evaluating a natural transformation $\alpha$ at $\alpha_x(1_x)$. So, in this case, natural transformations $\delta, \gamma : \hom_C(x,-) \Rightarrow F$ are equal if and only if $\delta_x(1_x) = \gamma_x(1_x)$. The restrictive part here is that the equality ought to happen in the component of $x$, but at the same time equality of the natural transformations can be concluded just from evaluating at a single element. 
