[proofplan]
We construct the bijection explicitly. Evaluation at the identity morphism gives a function from natural transformations to elements of $F(A)$, and an element $a\in F(A)$ gives a natural transformation by pushing $a$ forward along each morphism $u:A\to X$. Functoriality proves that this family is natural, and the naturality square for an arbitrary natural transformation evaluated at $u:A\to X$ proves that the two constructions are inverse.
[/proofplan]
[step:Define the evaluation map at the identity of $A$]
Since $\mathcal C$ is locally small, $\operatorname{Hom}_{\mathcal C}(A,X)$ is a set for every object $X$ of $\mathcal C$, so $\operatorname{Hom}_{\mathcal C}(A,-):\mathcal C\to \mathbf{Set}$ is a well-defined functor. For a natural transformation
\begin{align*}
\eta:\operatorname{Hom}_{\mathcal C}(A,-)\Rightarrow F,
\end{align*}
its component at $A$ is a function
\begin{align*}
\eta_A:\operatorname{Hom}_{\mathcal C}(A,A)\to F(A).
\end{align*}
Therefore $\eta_A(\operatorname{id}_A)\in F(A)$ is defined. Set
\begin{align*}
\Phi_{A,F}:\operatorname{Nat}(\operatorname{Hom}_{\mathcal C}(A,-),F) &\to F(A) \\
\eta &\mapsto \eta_A(\operatorname{id}_A).
\end{align*}
[/step]
[step:Construct a natural transformation from an element of $F(A)$]
Fix an element $a\in F(A)$. For every object $X$ of $\mathcal C$, define a function
\begin{align*}
\eta^a_X:\operatorname{Hom}_{\mathcal C}(A,X) &\to F(X) \\
u &\mapsto F(u)(a).
\end{align*}
We verify that the family $\eta^a=(\eta^a_X)_X$ is a natural transformation
\begin{align*}
\eta^a:\operatorname{Hom}_{\mathcal C}(A,-)\Rightarrow F.
\end{align*}
Let $f:X\to Y$ be a morphism in $\mathcal C$. Naturality requires the commutativity of
\begin{align*}
F(f)\circ \eta^a_X
=
\eta^a_Y\circ \operatorname{Hom}_{\mathcal C}(A,f),
\end{align*}
as functions from $\operatorname{Hom}_{\mathcal C}(A,X)$ to $F(Y)$. Let $u\in \operatorname{Hom}_{\mathcal C}(A,X)$. The left-hand side sends $u$ to
\begin{align*}
(F(f)\circ \eta^a_X)(u)
=
F(f)(F(u)(a)).
\end{align*}
Since $F$ is a covariant functor, $F(f)(F(u)(a))=(F(f)\circ F(u))(a)=F(f\circ u)(a)$. The right-hand side sends $u$ to
\begin{align*}
(\eta^a_Y\circ \operatorname{Hom}_{\mathcal C}(A,f))(u)
=
\eta^a_Y(f\circ u)
=
F(f\circ u)(a).
\end{align*}
Thus both sides agree on every $u\in \operatorname{Hom}_{\mathcal C}(A,X)$, so $\eta^a$ is natural.
[guided]
Fix $a\in F(A)$. The intended natural transformation should send each morphism $u:A\to X$ to the result of transporting $a$ along $u$ by the functor $F$. Thus, for each object $X$ of $\mathcal C$, define
\begin{align*}
\eta^a_X:\operatorname{Hom}_{\mathcal C}(A,X) &\to F(X) \\
u &\mapsto F(u)(a).
\end{align*}
This is a well-defined function because $u:A\to X$ implies $F(u):F(A)\to F(X)$, and $a\in F(A)$.
We now check naturality. Let $f:X\to Y$ be a morphism in $\mathcal C$. We must prove that the square comparing the two functions
\begin{align*}
\operatorname{Hom}_{\mathcal C}(A,X)\to F(Y)
\end{align*}
commutes, namely
\begin{align*}
F(f)\circ \eta^a_X
=
\eta^a_Y\circ \operatorname{Hom}_{\mathcal C}(A,f).
\end{align*}
Take an arbitrary morphism $u\in \operatorname{Hom}_{\mathcal C}(A,X)$. The upper-then-right route sends $u$ first to $F(u)(a)\in F(X)$ and then applies $F(f)$:
\begin{align*}
(F(f)\circ \eta^a_X)(u)
=
F(f)(F(u)(a)).
\end{align*}
Because $F$ is covariant, it preserves composition, so
\begin{align*}
F(f)(F(u)(a))
=
(F(f)\circ F(u))(a)
=
F(f\circ u)(a).
\end{align*}
The left-then-bottom route first sends $u$ to $f\circ u\in \operatorname{Hom}_{\mathcal C}(A,Y)$ and then applies $\eta^a_Y$:
\begin{align*}
(\eta^a_Y\circ \operatorname{Hom}_{\mathcal C}(A,f))(u)
=
\eta^a_Y(f\circ u)
=
F(f\circ u)(a).
\end{align*}
Both routes give the same element of $F(Y)$ for every $u:A\to X$, so the naturality square commutes. Hence $\eta^a:\operatorname{Hom}_{\mathcal C}(A,-)\Rightarrow F$ is a natural transformation.
[/guided]
[/step]
[step:Show that evaluation followed by reconstruction returns the original element]
Let $a\in F(A)$, and let $\eta^a$ be the natural transformation constructed above. Then
\begin{align*}
\Phi_{A,F}(\eta^a)
=
\eta^a_A(\operatorname{id}_A)
=
F(\operatorname{id}_A)(a)
=
\operatorname{id}_{F(A)}(a)
=
a,
\end{align*}
where the third equality uses that $F$ preserves identity morphisms. Thus $\Phi_{A,F}$ sends the reconstruction from $a$ back to $a$.
[/step]
[step:Show that reconstruction from the evaluated element returns the original transformation]
Let
\begin{align*}
\eta:\operatorname{Hom}_{\mathcal C}(A,-)\Rightarrow F
\end{align*}
be a natural transformation, and define
\begin{align*}
a:=\eta_A(\operatorname{id}_A)\in F(A).
\end{align*}
We prove that $\eta=\eta^a$. Let $X$ be an object of $\mathcal C$ and let $u\in \operatorname{Hom}_{\mathcal C}(A,X)$. Naturality of $\eta$ for the morphism $u:A\to X$ gives the commutative identity
\begin{align*}
F(u)\circ \eta_A
=
\eta_X\circ \operatorname{Hom}_{\mathcal C}(A,u),
\end{align*}
as functions from $\operatorname{Hom}_{\mathcal C}(A,A)$ to $F(X)$. Evaluating both sides at $\operatorname{id}_A$ gives
\begin{align*}
F(u)(\eta_A(\operatorname{id}_A))
=
\eta_X(\operatorname{Hom}_{\mathcal C}(A,u)(\operatorname{id}_A)).
\end{align*}
Since $\operatorname{Hom}_{\mathcal C}(A,u)(\operatorname{id}_A)=u\circ \operatorname{id}_A=u$, this becomes
\begin{align*}
F(u)(a)=\eta_X(u).
\end{align*}
By definition of $\eta^a_X$, the left-hand side is $\eta^a_X(u)$. Hence $\eta_X(u)=\eta^a_X(u)$ for every object $X$ and every $u:A\to X$, so $\eta=\eta^a$.
[guided]
Let $\eta:\operatorname{Hom}_{\mathcal C}(A,-)\Rightarrow F$ be any natural transformation. We want to show that knowing only the element
\begin{align*}
a:=\eta_A(\operatorname{id}_A)\in F(A)
\end{align*}
determines every component $\eta_X$ and every value $\eta_X(u)$.
Fix an object $X$ of $\mathcal C$ and a morphism $u:A\to X$. Apply naturality of $\eta$ to this particular morphism $u$. The naturality square says that the two functions from $\operatorname{Hom}_{\mathcal C}(A,A)$ to $F(X)$ agree:
\begin{align*}
F(u)\circ \eta_A
=
\eta_X\circ \operatorname{Hom}_{\mathcal C}(A,u).
\end{align*}
Now evaluate this identity at the identity morphism $\operatorname{id}_A\in \operatorname{Hom}_{\mathcal C}(A,A)$. The left-hand side becomes
\begin{align*}
(F(u)\circ \eta_A)(\operatorname{id}_A)
=
F(u)(\eta_A(\operatorname{id}_A))
=
F(u)(a).
\end{align*}
The right-hand side becomes
\begin{align*}
(\eta_X\circ \operatorname{Hom}_{\mathcal C}(A,u))(\operatorname{id}_A)
=
\eta_X(u\circ \operatorname{id}_A)
=
\eta_X(u).
\end{align*}
Therefore
\begin{align*}
\eta_X(u)=F(u)(a).
\end{align*}
But the natural transformation reconstructed from $a$ was defined by
\begin{align*}
\eta^a_X(u)=F(u)(a).
\end{align*}
Thus $\eta_X(u)=\eta^a_X(u)$ for every object $X$ and every morphism $u:A\to X$. Since natural transformations are equal exactly when all their components are equal as functions, $\eta=\eta^a$.
[/guided]
[/step]
[step:Conclude that the correspondence is a bijection]
The previous two steps show that the assignment
\begin{align*}
a\in F(A)\mapsto \eta^a\in \operatorname{Nat}(\operatorname{Hom}_{\mathcal C}(A,-),F)
\end{align*}
is both a left inverse and a right inverse to $\Phi_{A,F}$. Therefore $\Phi_{A,F}$ is a bijection, with inverse $a\mapsto \eta^a$. This proves the stated correspondence
\begin{align*}
\operatorname{Nat}(\operatorname{Hom}_{\mathcal C}(A,-),F)\cong F(A),
\end{align*}
and identifies the element corresponding to $\eta$ as $\eta_A(\operatorname{id}_A)$.
[/step]
