Isomorphism of hom sets implies objects are isomorphic Suppose we have 
$$Hom(x_0, y) \simeq Hom(x_1, y)\ \forall y \in \mathcal{C}$$
Does it follow that $x_0 \simeq x_1$ in $\mathcal{C}$? 
Now suppose that $F, G: \mathcal{C} \rightarrow \mathcal{D}$ are two functors such that $Fx \simeq Gx\ \forall x\in \mathcal{C}$. Does it follow that any natural transformation $\epsilon: F \Rightarrow G$ is a natural isomorphism? 
 A: The answer is no for both questions.
For the first one, define a category $\mathcal{C}$ with two objects $x_0,x_1$, the set of morphisms $x\to y$ is $\{0,1,\dots\}$ if $x=y$ and $\{1,2,\dots\}$ otherwise, and composition is just addition of natural numbers. Then for all $y$ there is a bijection between $\operatorname{Hom}(x_0,y)$ and $\operatorname{Hom}(x_1,y)$ (since both are infinite and countable), but there is no isomorphism $x_0\to x_1$.
For a less artificial counterexampe, take the category of finite-dimensional vector spaces over $\mathbb{R}$; then $\operatorname{Hom}(x,y)$ is in bijection with $\mathbb{R}^{\dim(y)\times \dim(x)}$, so it has the same cardinality as $\mathbb{R}$ as long as neither $x$ nor $y$ is the zero vector space. In particular, for all non-zero vector spaces $x_0,x_1$, there is a bijection between $\operatorname{Hom}(x_0,y)$ and $\operatorname{Hom}(x_1,y)$ for all $y$.
For the second question, take any counterexample to the first one, and take $\mathcal{D}=\mathbf{Set}$ and $F,G$ the functors represented by $x_0$ and $x_1$ respectively. Then there exist morphisms $x_0\to x_1$ and $x_1\to x_0$, and thus natural transformations $G\Rightarrow F$ and $F\Rightarrow G$, but $F$ and $G$ are not isomorphic since $x_0$ and $x_1$ aren't. Note that even if $F$ and $G$ were isomorphic, there could still be natural transformations between them that are not isomorphisms.
You can also find plenty of counterexamples to your second question in this MO thread.
