[proofplan]
Use initiality twice to obtain the unique morphisms $u: I \to I'$ and $v: I' \to I$. Then compare the composites $v \circ u$ and $u \circ v$ with the corresponding identity morphisms. Since initiality also makes each endomorphism set of an initial object a singleton, these composites must be identities, so $u$ and $v$ are inverse isomorphisms. Finally, uniqueness of the isomorphism $I \to I'$ follows from the fact that there is only one morphism from $I$ to $I'$.
[/proofplan]
[step:Construct the two morphisms supplied by initiality]
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*}
u: 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*}
v: I' \to I.
\end{align*}
[/step]
[step:Show that the two composites are identity morphisms]
The composite
\begin{align*}
v \circ u: I \to I
\end{align*}
is an element of $\operatorname{Hom}_{\mathcal C}(I,I)$. The identity morphism
\begin{align*}
\operatorname{id}_I: I \to I
\end{align*}
is also an element of $\operatorname{Hom}_{\mathcal C}(I,I)$. Since $I$ is initial, $\operatorname{Hom}_{\mathcal C}(I,I)$ contains exactly one morphism. Therefore
\begin{align*}
v \circ u = \operatorname{id}_I.
\end{align*}
Similarly, the composite
\begin{align*}
u \circ v: I' \to I'
\end{align*}
and the identity morphism
\begin{align*}
\operatorname{id}_{I'}: I' \to I'
\end{align*}
both belong to $\operatorname{Hom}_{\mathcal C}(I',I')$. Since $I'$ is initial, this hom-set contains exactly one morphism. Hence
\begin{align*}
u \circ v = \operatorname{id}_{I'}.
\end{align*}
[guided]
We must prove that $u$ is an isomorphism, so we need a morphism in the opposite direction and then we need to check the two inverse equations. Initiality supplies the opposite morphism $v: I' \to I$.
Now consider the composite $v \circ u$. Its domain and codomain are both $I$, so
\begin{align*}
v \circ u: I \to I.
\end{align*}
The identity morphism on $I$ is another morphism with the same domain and codomain:
\begin{align*}
\operatorname{id}_I: I \to I.
\end{align*}
Because $I$ is initial, there is exactly one morphism from $I$ to any object of $\mathcal C$, and in particular exactly one morphism from $I$ to itself. Thus the two morphisms $v \circ u$ and $\operatorname{id}_I$ must be equal:
\begin{align*}
v \circ u = \operatorname{id}_I.
\end{align*}
The same argument applies to $I'$. The composite $u \circ v$ and the identity morphism $\operatorname{id}_{I'}$ are both morphisms from $I'$ to $I'$:
\begin{align*}
u \circ v: I' \to I',
\qquad
\operatorname{id}_{I'}: I' \to I'.
\end{align*}
Since $I'$ is initial, $\operatorname{Hom}_{\mathcal C}(I',I')$ contains exactly one morphism. Therefore
\begin{align*}
u \circ v = \operatorname{id}_{I'}.
\end{align*}
[/guided]
[/step]
[step:Conclude that the morphism is the unique isomorphism]
The equalities
\begin{align*}
v \circ u = \operatorname{id}_I,
\qquad
u \circ v = \operatorname{id}_{I'}
\end{align*}
show that $v$ is a two-sided inverse of $u$. Hence $u: I \to I'$ is an isomorphism in $\mathcal C$.
It remains to prove uniqueness. Let
\begin{align*}
w: I \to I'
\end{align*}
be any isomorphism in $\mathcal C$. Since $w$ is a morphism from $I$ to $I'$ and $\operatorname{Hom}_{\mathcal C}(I,I')$ contains exactly one morphism, we have
\begin{align*}
w = u.
\end{align*}
Therefore $u$ is the unique isomorphism from $I$ to $I'$.
[/step]
