Natural isomorphism of hom functors imply isomorphism of objects

Yes. The Yoneda embedding $$y:\operatorname{Hom}_{\mathcal{C}}(A,B)\to\operatorname{Hom}_{\mathbf{Set}^{\mathcal{C}^{op}}}(\mathcal{C}(-,A),\mathcal{C}(-,B)),\quad y(X) = \mathcal{C}(-,X)$$ is fully faithful, and therefore reflects isomorphisms. Hence, given any natural isomorphism $\eta:\mathcal{C}(-,A)\cong \mathcal{C}(-,B)$, there exists an isomorphism $f:A\to B$ such that $y(f)=\eta$.

Yes (as the other answer points out). But here's a proof that does not invoke Yoneda.

It is given that $\mathcal C(-, A) \cong \mathcal C(-, B)$ naturally in $-$.

Plugging in $A$: $\mathcal C(A, A) \cong \mathcal C(A, B)$ naturally in $A$.

But $1_A$ is an element of $\mathcal C(A, A)$, and is an isomorphism, and the natural isomorphism from $\mathcal C(A, A)$ to $\mathcal C(A, B)$ must map this to an isomorphism in $\mathcal C(A, B)$, which means $A \cong B$.

To prove that last claim, let $\alpha \colon \mathcal C(-, A) \Rightarrow \mathcal C(-, B)$ be a natural isomorphism, and $\alpha^{-1} \colon \mathcal C(-,B) \to \mathcal C(-, A)$ its inverse. Let $f = \alpha_A(1_A) \in \mathcal C(A, B)$. The naturality square is then as given below. $\require{AMScd}$ \begin{CD} \mathcal C(A, A) @>{\alpha_A}>> \mathcal C(A, B)\\ @A(-\circ f)AA @AA(- \circ f)A \\ \mathcal C(B,A) @>>{\alpha_B}> \mathcal C(B, B) \end{CD} Here $- \circ f$ is the function $\mathcal C(f, A)$ (and also $\mathcal C(f, B)$) whose action is to take a morphism from $\mathcal C(A,A)$ (or $\mathcal C(A,B)$) and compose it with $f$.

We must find an inverse $g \colon B \to A$ for $f$, and an obvious choice is $g = \alpha_B^{-1}(1_B) \in \mathcal C(B, A)$. So now we just need to verify that these are inverses.

First, observe from the naturality square that \begin{equation*} \alpha_A \circ (- \circ f) = (- \circ f) \circ \alpha_B = (\alpha_B(-)) \circ f. \end{equation*} Applying these functions to $g$ (which is an element of $\mathcal C(B, A)$), we get \begin{align*} \alpha_A(g \circ f) &= \alpha_B(g) \circ f\\ &= 1_B \circ f\\ &= f. \end{align*} Then $g \circ f = \alpha_A^{-1}(f) = 1_A$.

Similarly, $f \circ g = 1_B$.