[proofplan] We compare the two representing isomorphisms by forming the natural isomorphism $\tau:=\beta^{-1}\circ\alpha$ from $\operatorname{Hom}_{\mathcal C}(-,A)$ to $\operatorname{Hom}_{\mathcal C}(-,B)$. Naturality forces $\tau$ to be postcomposition by the single morphism $f:=\tau_A(\operatorname{id}_A):A\to B$. Applying the same construction to $\tau^{-1}$ produces a morphism $g:B\to A$, and the two natural transformations being inverse implies $g\circ f=\operatorname{id}_A$ and $f\circ g=\operatorname{id}_B$. Finally, the compatibility equation determines $f$ by evaluating at $X=A$ and $u=\operatorname{id}_A$. [/proofplan] [step:Compare the two representing isomorphisms by a natural isomorphism of Hom functors] Define a natural transformation \begin{align*} \tau:\operatorname{Hom}_{\mathcal C}(-,A)&\longrightarrow \operatorname{Hom}_{\mathcal C}(-,B) \end{align*} by \begin{align*} \tau_X:=\beta_X^{-1}\circ \alpha_X \end{align*} for each object $X\in \operatorname{Ob}(\mathcal C)$. Since $\alpha$ and $\beta$ are natural isomorphisms, each component $\tau_X$ is a bijection, and $\tau$ is a natural isomorphism. Thus, for every morphism $r:Y\to X$ in $\mathcal C$, the naturality square says that for every $u:X\to A$, \begin{align*} \tau_Y(u\circ r)=\tau_X(u)\circ r. \end{align*} [/step] [step:Recover the morphism $A\to B$ from the image of the identity] Define the morphism \begin{align*} f:A&\longrightarrow B \end{align*} by \begin{align*} f:=\tau_A(\operatorname{id}_A). \end{align*} We claim that $\tau$ is postcomposition by $f$. Let $X\in \operatorname{Ob}(\mathcal C)$ and let $u:X\to A$ be a morphism. Apply the naturality identity for $\tau$ to the morphism $u:X\to A$ and to $\operatorname{id}_A\in \operatorname{Hom}_{\mathcal C}(A,A)$. This gives \begin{align*} \tau_X(u) &=\tau_X(\operatorname{id}_A\circ u)\\ &=\tau_A(\operatorname{id}_A)\circ u\\ &=f\circ u. \end{align*} Therefore, for every object $X$ and every morphism $u:X\to A$, \begin{align*} \beta_X^{-1}(\alpha_X(u))=\tau_X(u)=f\circ u. \end{align*} Applying $\beta_X$ to both sides yields the desired compatibility relation \begin{align*} \beta_X(f\circ u)=\alpha_X(u). \end{align*} [guided] The natural isomorphism $\tau$ contains all the information comparing the two representing objects. To extract an actual morphism $A\to B$, we evaluate $\tau$ at the object $A$ and at the distinguished element $\operatorname{id}_A\in \operatorname{Hom}_{\mathcal C}(A,A)$. Define \begin{align*} f:A&\longrightarrow B \end{align*} by \begin{align*} f:=\tau_A(\operatorname{id}_A). \end{align*} Now we show that this single morphism controls every component of $\tau$. Let $X\in \operatorname{Ob}(\mathcal C)$ and let $u:X\to A$ be a morphism. Since $\operatorname{Hom}_{\mathcal C}(-,A)$ and $\operatorname{Hom}_{\mathcal C}(-,B)$ are contravariant Hom functors, the morphism $u:X\to A$ induces maps by precomposition: \begin{align*} \operatorname{Hom}_{\mathcal C}(A,A)&\longrightarrow \operatorname{Hom}_{\mathcal C}(X,A),\\ h&\longmapsto h\circ u, \end{align*} and \begin{align*} \operatorname{Hom}_{\mathcal C}(A,B)&\longrightarrow \operatorname{Hom}_{\mathcal C}(X,B),\\ k&\longmapsto k\circ u. \end{align*} Naturality of $\tau$ with respect to $u:X\to A$ therefore gives \begin{align*} \tau_X(\operatorname{id}_A\circ u)=\tau_A(\operatorname{id}_A)\circ u. \end{align*} Since $\operatorname{id}_A\circ u=u$ and $\tau_A(\operatorname{id}_A)=f$, this becomes \begin{align*} \tau_X(u)=f\circ u. \end{align*} Because $\tau_X=\beta_X^{-1}\circ\alpha_X$, we have \begin{align*} \beta_X^{-1}(\alpha_X(u))=f\circ u. \end{align*} Applying the bijection $\beta_X$ to both sides gives \begin{align*} \beta_X(f\circ u)=\alpha_X(u). \end{align*} This is exactly the compatibility condition in the statement. [/guided] [/step] [step:Construct the inverse morphism from the inverse natural isomorphism] Let \begin{align*} \sigma:\operatorname{Hom}_{\mathcal C}(-,B)&\longrightarrow \operatorname{Hom}_{\mathcal C}(-,A) \end{align*} be the inverse natural isomorphism $\sigma:=\tau^{-1}$. Define \begin{align*} g:B&\longrightarrow A \end{align*} by \begin{align*} g:=\sigma_B(\operatorname{id}_B). \end{align*} The same naturality argument applied to $\sigma$ shows that, for every object $X\in \operatorname{Ob}(\mathcal C)$ and every morphism $v:X\to B$, \begin{align*} \sigma_X(v)=g\circ v. \end{align*} Since $\sigma\circ\tau=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(-,A)}$, evaluating at $A$ and at $\operatorname{id}_A$ gives \begin{align*} \operatorname{id}_A &=(\sigma_A\circ\tau_A)(\operatorname{id}_A)\\ &=\sigma_A(f)\\ &=g\circ f. \end{align*} Since $\tau\circ\sigma=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(-,B)}$, evaluating at $B$ and at $\operatorname{id}_B$ gives \begin{align*} \operatorname{id}_B &=(\tau_B\circ\sigma_B)(\operatorname{id}_B)\\ &=\tau_B(g)\\ &=f\circ g. \end{align*} Thus $g$ is a two-sided inverse for $f$, so $f:A\to B$ is an isomorphism in $\mathcal C$. [/step] [step:Prove uniqueness among compatible morphisms] Let $q:A\to B$ be a morphism satisfying the same compatibility condition: for every object $X\in \operatorname{Ob}(\mathcal C)$ and every morphism $u:X\to A$, \begin{align*} \beta_X(q\circ u)=\alpha_X(u). \end{align*} Taking $X=A$ and $u=\operatorname{id}_A$, we obtain \begin{align*} \beta_A(q)=\alpha_A(\operatorname{id}_A). \end{align*} For the morphism $f$ constructed above, the same compatibility condition gives \begin{align*} \beta_A(f)=\alpha_A(\operatorname{id}_A). \end{align*} Hence \begin{align*} \beta_A(q)=\beta_A(f). \end{align*} Since $\beta_A:\operatorname{Hom}_{\mathcal C}(A,B)\to F(A)$ is a bijection, it follows that $q=f$. Therefore the compatible isomorphism $A\to B$ is unique. [/step]