Prove that these two objects are isomorphic So we have a category $C$ and objects $X$ and $Y$. Given that Mor$_C(-,X)$ and Mor$_C(-,Y)$ are isomorphic functors, then prove that $X\cong Y$ (as objects in $C$). 
So by the condition I know that there exists a natural transformation $\eta$ such that for each object $A\in C$ we obtain the map $\eta_A$ such that $\eta_A:$Mor$_C(A,X)\rightarrow$ Mor$_C(A,Y)$. Further, we have that each $\eta_A$ is an isomorphism. My doubts comes on what $\eta$ does to Mor$_C(A,X)$, does it take an arrow from $A$ to $X$ and it makes it now point to $Y$? My intuition tells me that if we let $A=X$, then we have two arrows (call them $f$ and $g$) in Mor$_C(X,X)$ such that $fg=1_X$, and somehow I feel that $\eta_X$ should take $f$ and $g$ to arrows that do the same thing. 
Could someone please tell me whether my intuition is correct and perhaps show me how to do this with rigor?
Thanks.
 A: $\newcommand{\mc}{\mathcal}$Yes, suppose we have the natural isomorphism $$\eta : \mc C(-,A) \to \mc C(-,B)$$ then we have, for arbitrary $f : B \to A$ the naturality condition: $\mc C(f,B) \circ \eta_A = \eta_B \circ \mc C(f,A)$ which, applying both sides to $1_A$ and by definition of the Hom-functor, gives $f \circ \eta_A(1_A) = \eta_B(f)$. Setting $f = \eta^{-1}_B(1_B)$ we find that $\eta_A(1_A) \circ \eta^{-1}_B(1_B) = 1_B$ so (after a similar argument to show that it is also the left inverse) $\eta_A(1_A) : A \to B$ is an isomorphism.

Recall that $\mc C(-,C) : \mc C^\text{op} \to \text{Set}$ is a contravariant functor. When $f : X \to Y$ is a morphism in $\mc C$ we have $\mc C(f,C) : \mc C(Y,C) \to C(X,C)$ defined, for $g \in C(Y,C)$ by $\mc C(f,C)(g) = g \circ f$ because $X \xrightarrow{f} Y \xrightarrow{g} C$.
Let us see the full calculation with this $$
\begin{eqnarray}{}
\mc C(f,B) \circ \eta_A &=& \eta_B \circ \mc C(f,A) \\
\mc C(f,B)(\eta_A(1_A)) &=& \eta_B(\mc C(f,A)(1_A)) \\
\mc \eta_A(1_A) \circ f &=& \eta_B(1_A \circ f) \\
\mc \eta_A(1_A) \circ f &=& \eta_B(f) \\
\end{eqnarray}$$
