[proofplan] We prove the two constructions are inverse in content. Starting from a natural isomorphism $\Phi:H_A\cong F$, we take the element represented by the identity morphism, namely $u=\Phi_A(\operatorname{id}_A)$, and use naturality to show that every component $\Phi_X$ is the map $f\mapsto F(f)(u)$. Since each $\Phi_X$ is bijective, this gives the required existence and uniqueness. Conversely, starting from such an element $u$, we define $\Phi_X(f)=F(f)(u)$; the universal property gives bijectivity of each component, and functoriality of $F$ gives naturality. [/proofplan] [step:Extract the universal element from a natural representation] Assume first that $\Phi:H_A\cong F$ is a natural isomorphism. For each object $X$ of $\mathcal C$, write \begin{align*} \Phi_X:\operatorname{Hom}_{\mathcal C}(X,A)\to F(X) \end{align*} for the component of $\Phi$ at $X$. Define \begin{align*} u:=\Phi_A(\operatorname{id}_A)\in F(A), \end{align*} where $\operatorname{id}_A:A\to A$ is the identity morphism of $A$. Let $X$ be an object of $\mathcal C$ and let $f:X\to A$ be a morphism. Naturality of $\Phi$ with respect to $f:X\to A$ says that \begin{align*} F(f)\circ \Phi_A=\Phi_X\circ H_A(f). \end{align*} Evaluating this equality at $\operatorname{id}_A\in \operatorname{Hom}_{\mathcal C}(A,A)$ gives \begin{align*} F(f)(u) &=F(f)(\Phi_A(\operatorname{id}_A)) \\ &=\Phi_X(H_A(f)(\operatorname{id}_A)) \\ &=\Phi_X(\operatorname{id}_A\circ f) \\ &=\Phi_X(f). \end{align*} Thus, for every morphism $f:X\to A$, \begin{align*} \Phi_X(f)=F(f)(u). \end{align*} [/step] [step:Use bijectivity of the representation to prove the universal property] Let $X$ be an object of $\mathcal C$ and let $s\in F(X)$. Since $\Phi$ is a natural isomorphism, the component \begin{align*} \Phi_X:\operatorname{Hom}_{\mathcal C}(X,A)\to F(X) \end{align*} is a bijection. Therefore there exists a unique morphism $f:X\to A$ such that \begin{align*} \Phi_X(f)=s. \end{align*} By the identity proved in the previous step, $\Phi_X(f)=F(f)(u)$, so this unique morphism is precisely the unique morphism satisfying \begin{align*} F(f)(u)=s. \end{align*} Hence $u\in F(A)$ has the stated universal property. [/step] [step:Construct a natural transformation from a universal element] Conversely, assume that an element $u\in F(A)$ has the stated universal property. For each object $X$ of $\mathcal C$, define a function \begin{align*} \Phi_X:\operatorname{Hom}_{\mathcal C}(X,A)&\to F(X),\\ f&\mapsto F(f)(u). \end{align*} This is well-defined because every morphism $f:X\to A$ is sent by the contravariant functor $F$ to a function \begin{align*} F(f):F(A)\to F(X), \end{align*} so $F(f)(u)$ is an element of $F(X)$. For each object $X$ of $\mathcal C$, the universal property of $u$ states exactly that for every $s\in F(X)$ there exists a unique $f\in \operatorname{Hom}_{\mathcal C}(X,A)$ with $\Phi_X(f)=s$. Hence each function $\Phi_X$ is bijective. [/step] [step:Verify naturality of the constructed bijections] Let $g:Y\to X$ be a morphism in $\mathcal C$. We must prove that the square \begin{align*} F(g)\circ \Phi_X=\Phi_Y\circ H_A(g) \end{align*} commutes as functions from $\operatorname{Hom}_{\mathcal C}(X,A)$ to $F(Y)$. Let $f:X\to A$ be a morphism. By the definition of $\Phi_X$, the definition of $H_A(g)$, and functoriality of the contravariant functor $F:\mathcal C^{\mathrm{op}}\to \mathbf{Set}$, we have \begin{align*} (F(g)\circ \Phi_X)(f) &=F(g)(\Phi_X(f)) \\ &=F(g)(F(f)(u)) \\ &=(F(g)\circ F(f))(u) \\ &=F(f\circ g)(u) \\ &=\Phi_Y(f\circ g) \\ &=\Phi_Y(H_A(g)(f)) \\ &=(\Phi_Y\circ H_A(g))(f). \end{align*} Since this equality holds for every $f\in \operatorname{Hom}_{\mathcal C}(X,A)$, the naturality square commutes. Therefore the family $(\Phi_X)_X$ is a natural transformation $\Phi:H_A\to F$. Each component $\Phi_X$ is bijective by the previous step, so $\Phi$ is a natural isomorphism. This proves that the universal element $u\in F(A)$ determines a representation $H_A\cong F$, completing the equivalence. [/step]