[proofplan]
We prove the bijection directly. A morphism $r:B \to A$ defines a natural transformation by precomposition, and naturality follows from associativity of composition in $\mathcal C$. Conversely, every natural transformation is determined by its value at $\operatorname{id}_A \in \operatorname{Hom}_{\mathcal C}(A,A)$. These two constructions are inverse to each other.
[/proofplan]
[step:Define the natural transformation determined by a morphism $r:B \to A$]
Let $r \in \operatorname{Hom}_{\mathcal C}(B,A)$. For each object $X \in \operatorname{Ob}(\mathcal C)$, define a function
\begin{align*}
\tau_X^r:\operatorname{Hom}_{\mathcal C}(A,X) &\to \operatorname{Hom}_{\mathcal C}(B,X) \\
u &\mapsto u \circ r.
\end{align*}
This function is well-defined because $u:A \to X$ and $r:B \to A$, so the composite $u \circ r:B \to X$ is a morphism of $\mathcal C$.
We verify that the family $\tau^r = (\tau_X^r)_{X \in \operatorname{Ob}(\mathcal C)}$ is a natural transformation
\begin{align*}
\tau^r:\operatorname{Hom}_{\mathcal C}(A,-) \Rightarrow \operatorname{Hom}_{\mathcal C}(B,-).
\end{align*}
Let $f:X \to Y$ be a morphism in $\mathcal C$, and let $u \in \operatorname{Hom}_{\mathcal C}(A,X)$. The covariant Hom functors act on $f$ by postcomposition:
\begin{align*}
\operatorname{Hom}_{\mathcal C}(A,f)(u) &= f \circ u, \\
\operatorname{Hom}_{\mathcal C}(B,f)(u \circ r) &= f \circ u \circ r.
\end{align*}
Therefore, using associativity of composition in $\mathcal C$,
\begin{align*}
\operatorname{Hom}_{\mathcal C}(B,f)\bigl(\tau_X^r(u)\bigr)
&= \operatorname{Hom}_{\mathcal C}(B,f)(u \circ r) \\
&= f \circ (u \circ r) \\
&= (f \circ u) \circ r \\
&= \tau_Y^r(f \circ u) \\
&= \tau_Y^r\bigl(\operatorname{Hom}_{\mathcal C}(A,f)(u)\bigr).
\end{align*}
Since this holds for every $u \in \operatorname{Hom}_{\mathcal C}(A,X)$, the naturality square for $f$ commutes. Hence $\tau^r$ is a natural transformation.
[guided]
Fix a morphism $r:B \to A$. The intended transformation sends every arrow out of $A$ to an arrow out of $B$ by precomposing with $r$. Thus, for each object $X$, we define
\begin{align*}
\tau_X^r:\operatorname{Hom}_{\mathcal C}(A,X) &\to \operatorname{Hom}_{\mathcal C}(B,X) \\
u &\mapsto u \circ r.
\end{align*}
The domain and codomain match: if $u:A \to X$ and $r:B \to A$, then $u \circ r:B \to X$.
Now we check naturality. Let $f:X \to Y$ be a morphism in $\mathcal C$. Naturality requires that applying $\tau_X^r$ first and then postcomposing with $f$ gives the same result as postcomposing with $f$ first and then applying $\tau_Y^r$. For $u:A \to X$, the first route gives
\begin{align*}
\operatorname{Hom}_{\mathcal C}(B,f)\bigl(\tau_X^r(u)\bigr)
&= \operatorname{Hom}_{\mathcal C}(B,f)(u \circ r) \\
&= f \circ (u \circ r).
\end{align*}
The second route gives
\begin{align*}
\tau_Y^r\bigl(\operatorname{Hom}_{\mathcal C}(A,f)(u)\bigr)
&= \tau_Y^r(f \circ u) \\
&= (f \circ u) \circ r.
\end{align*}
Associativity of composition in the category $\mathcal C$ gives
\begin{align*}
f \circ (u \circ r) = (f \circ u) \circ r.
\end{align*}
Thus the two routes agree for every $u \in \operatorname{Hom}_{\mathcal C}(A,X)$, so the square commutes for every morphism $f:X \to Y$. Therefore $\tau^r$ is a natural transformation.
[/guided]
[/step]
[step:Recover a morphism $B \to A$ from a natural transformation]
Let
\begin{align*}
\eta:\operatorname{Hom}_{\mathcal C}(A,-) \Rightarrow \operatorname{Hom}_{\mathcal C}(B,-)
\end{align*}
be a natural transformation. Define
\begin{align*}
\Psi(\eta) := \eta_A(\operatorname{id}_A) \in \operatorname{Hom}_{\mathcal C}(B,A).
\end{align*}
This is well-defined because
\begin{align*}
\eta_A:\operatorname{Hom}_{\mathcal C}(A,A) \to \operatorname{Hom}_{\mathcal C}(B,A).
\end{align*}
Thus evaluation at $\operatorname{id}_A$ defines a function
\begin{align*}
\Psi:\operatorname{Nat}\!\left(\operatorname{Hom}_{\mathcal C}(A,-),\operatorname{Hom}_{\mathcal C}(B,-)\right) &\to \operatorname{Hom}_{\mathcal C}(B,A) \\
\eta &\mapsto \eta_A(\operatorname{id}_A).
\end{align*}
[/step]
[step:Show that evaluation at $\operatorname{id}_A$ determines every component]
Let
\begin{align*}
\eta:\operatorname{Hom}_{\mathcal C}(A,-) \Rightarrow \operatorname{Hom}_{\mathcal C}(B,-)
\end{align*}
be a natural transformation, and define $r := \Psi(\eta)=\eta_A(\operatorname{id}_A)$. Let $X \in \operatorname{Ob}(\mathcal C)$ and let $u \in \operatorname{Hom}_{\mathcal C}(A,X)$.
Apply naturality of $\eta$ to the morphism $u:A \to X$. The naturality identity gives
\begin{align*}
\operatorname{Hom}_{\mathcal C}(B,u)\bigl(\eta_A(\operatorname{id}_A)\bigr)
=
\eta_X\bigl(\operatorname{Hom}_{\mathcal C}(A,u)(\operatorname{id}_A)\bigr).
\end{align*}
Since
\begin{align*}
\operatorname{Hom}_{\mathcal C}(B,u)(r) &= u \circ r, \\
\operatorname{Hom}_{\mathcal C}(A,u)(\operatorname{id}_A) &= u \circ \operatorname{id}_A = u,
\end{align*}
we obtain
\begin{align*}
u \circ r = \eta_X(u).
\end{align*}
Therefore $\eta_X(u)=\tau_X^r(u)$ for every object $X$ and every $u:A \to X$, so $\eta=\tau^r$.
[guided]
The key point is that naturality with respect to the arrow $u:A \to X$ forces the value of $\eta_X$ on $u$ to be determined by the single element $\eta_A(\operatorname{id}_A)$. Define
\begin{align*}
r := \eta_A(\operatorname{id}_A) \in \operatorname{Hom}_{\mathcal C}(B,A).
\end{align*}
Now fix an object $X$ and a morphism $u:A \to X$.
Naturality of $\eta$ for the morphism $u:A \to X$ says that the following two ways of moving from $\operatorname{id}_A \in \operatorname{Hom}_{\mathcal C}(A,A)$ to an element of $\operatorname{Hom}_{\mathcal C}(B,X)$ agree:
\begin{align*}
\operatorname{Hom}_{\mathcal C}(B,u)\bigl(\eta_A(\operatorname{id}_A)\bigr)
=
\eta_X\bigl(\operatorname{Hom}_{\mathcal C}(A,u)(\operatorname{id}_A)\bigr).
\end{align*}
We compute both sides. Since $r=\eta_A(\operatorname{id}_A)$ and $\operatorname{Hom}_{\mathcal C}(B,u)$ postcomposes morphisms $B \to A$ with $u:A \to X$,
\begin{align*}
\operatorname{Hom}_{\mathcal C}(B,u)\bigl(\eta_A(\operatorname{id}_A)\bigr)
=
\operatorname{Hom}_{\mathcal C}(B,u)(r)
=
u \circ r.
\end{align*}
Also, since $\operatorname{Hom}_{\mathcal C}(A,u)$ postcomposes morphisms $A \to A$ with $u:A \to X$,
\begin{align*}
\operatorname{Hom}_{\mathcal C}(A,u)(\operatorname{id}_A)
=
u \circ \operatorname{id}_A
=
u.
\end{align*}
Substituting these two computations into the naturality identity gives
\begin{align*}
u \circ r = \eta_X(u).
\end{align*}
Thus every component $\eta_X$ is exactly precomposition by $r$, meaning $\eta=\tau^r$.
[/guided]
[/step]
[step:Verify that the two constructions are inverse bijections]
We now show that $\Phi$ and $\Psi$ are inverse functions.
First let $r \in \operatorname{Hom}_{\mathcal C}(B,A)$. Then
\begin{align*}
\Psi(\Phi(r))
&= \Psi(\tau^r) \\
&= \tau_A^r(\operatorname{id}_A) \\
&= \operatorname{id}_A \circ r \\
&= r.
\end{align*}
Thus $\Psi \circ \Phi$ is the identity function on $\operatorname{Hom}_{\mathcal C}(B,A)$.
Conversely, let
\begin{align*}
\eta \in \operatorname{Nat}\!\left(\operatorname{Hom}_{\mathcal C}(A,-),\operatorname{Hom}_{\mathcal C}(B,-)\right).
\end{align*}
By the previous step, if $r=\Psi(\eta)$, then $\eta=\tau^r=\Phi(r)$. Hence
\begin{align*}
\Phi(\Psi(\eta))=\eta.
\end{align*}
Therefore $\Phi \circ \Psi$ is the identity function on the natural transformation set.
The functions $\Phi$ and $\Psi$ are inverse to each other, so $\Phi$ is a bijection. This proves
\begin{align*}
\operatorname{Nat}\!\left(\operatorname{Hom}_{\mathcal C}(A,-),\operatorname{Hom}_{\mathcal C}(B,-)\right)
\cong
\operatorname{Hom}_{\mathcal C}(B,A),
\end{align*}
with $r:B \to A$ corresponding to the natural transformation whose component at $X$ sends $u:A \to X$ to $u \circ r:B \to X$.
[/step]
