[proofplan] We use initiality twice to construct the only possible morphisms $I \to I'$ and $I' \to I$. Then we compare their composites with the identity morphisms, using initiality again on the objects $I$ and $I'$. These identities show that the two morphisms are inverse isomorphisms, and the same uniqueness property shows that no other isomorphism $I \to I'$ can exist. [/proofplan] [step:Construct the unique morphisms between the two initial objects] Since $I$ is initial and $I'$ is an object of $\mathcal C$, the set $\operatorname{Hom}_{\mathcal C}(I, I')$ contains exactly one morphism. Denote this unique morphism by \begin{align*} f: I \to I'. \end{align*} Since $I'$ is initial and $I$ is an object of $\mathcal C$, the set $\operatorname{Hom}_{\mathcal C}(I', I)$ contains exactly one morphism. Denote this unique morphism by \begin{align*} g: I' \to I. \end{align*} [/step] [step:Show that the composite $g \circ f$ is the identity on $I$] The composite \begin{align*} g \circ f: I \to I \end{align*} is a morphism in $\operatorname{Hom}_{\mathcal C}(I, I)$. The identity morphism \begin{align*} \operatorname{id}_I: I \to I \end{align*} is also a morphism in $\operatorname{Hom}_{\mathcal C}(I, I)$. Because $I$ is initial, the set $\operatorname{Hom}_{\mathcal C}(I, I)$ contains exactly one morphism. Therefore \begin{align*} g \circ f = \operatorname{id}_I. \end{align*} [/step] [step:Show that the composite $f \circ g$ is the identity on $I'$] The composite \begin{align*} f \circ g: I' \to I' \end{align*} is a morphism in $\operatorname{Hom}_{\mathcal C}(I', I')$. The identity morphism \begin{align*} \operatorname{id}_{I'}: I' \to I' \end{align*} is also a morphism in $\operatorname{Hom}_{\mathcal C}(I', I')$. Because $I'$ is initial, the set $\operatorname{Hom}_{\mathcal C}(I', I')$ contains exactly one morphism. Therefore \begin{align*} f \circ g = \operatorname{id}_{I'}. \end{align*} [/step] [step:Conclude that $f$ is the unique isomorphism from $I$ to $I'$] The identities \begin{align*} g \circ f &= \operatorname{id}_I, \\ f \circ g &= \operatorname{id}_{I'} \end{align*} show that $f$ is an isomorphism with inverse $g$. It remains to prove uniqueness among isomorphisms. Let \begin{align*} h: I \to I' \end{align*} be any isomorphism in $\mathcal C$. Since $h$ is in $\operatorname{Hom}_{\mathcal C}(I, I')$ and $f$ is the unique morphism in $\operatorname{Hom}_{\mathcal C}(I, I')$, we have \begin{align*} h = f. \end{align*} Thus $f: I \to I'$ is the unique isomorphism from $I$ to $I'$. [/step]