Any object isomorphic to an initial object in a category is initial I am in trouble with a detail of the proof I've found, in particular with the fact that the isomorphism is unique but this property does not seem necessary.
The proof goes as follow.
Suppose I is initial and that I' is isomorphic to it. Let's say f is an iso with g as its inverse:

Given any other object X we want to show that there is a morphism $I' \rightarrow X$ and we want to show that this arrow is unique, so I' is initial.
Now, since I is initial there is a unique morphism $I \rightarrow X$, say s, and so we have at least a morphism $I' \rightarrow X$, that is $s \circ g$. To show that there are no other possible arrows $I' \rightarrow X$ we pick any arrow $t: I' \rightarrow X$ and we notice that:
$$
s = t \circ f \tag*{both arrows $I \rightarrow X$, $s$ is unique}
$$
$$
s \circ g = t \circ f \circ g \tag*{precomposition of g} 
$$
$$
s \circ g = t \tag*{$f \circ g = id$} 
$$
So any arrow $t: I' \rightarrow X$ is exactly like $s \circ g$, and that concludes the proof.
What bothers me is the fact that the proof does not use the uniqueness of the isomorphism between $I$ and $I'$. We've just proven that given any isomorphism $f: I \rightarrow I'$, $g: I' \rightarrow I$, any arrow $t: I' \rightarrow X$ is the same the unique arrow that goes from $I$ to $X$ precomposed with $g$. But we need that the iso $f, g$ is unique to have well behaviour: let's say there is another iso $f',g'$ between $I$ and $I'$, $t$ would be equal to two different things, $s \circ g$ and $s \circ g'$, because the same reasoning would apply.
That is just not possible, because $I$ is initial so the iso $f,g$ is unique, so it feels like this fact should be exploited, but the book says: "we did not need to use the fact that $I$ and $I'$ are uniquely isomorphic as that uniqueness follows from $I$ being initial".
I am confused.
 A: I feel that in order to highlight why the uniqueness of the isomorphism is not necessary, it's valuable to prove the analogous statement for coproducts.
Suppose $X$ and $Y$ are objects of some category, and suppose we have some isomorphism $f:X\sqcup Y\to Z$ from a choice of coproduct $X\sqcup Y$ to another object $Z$.
Then, $Z$ is also a coproduct of $X$ and $Y$!
Of course, the above statement is imprecise: a coproduct is not just an object, but an object equipped with some additional structure.
More precisely, the coproduct $X\sqcup Y$ comes with canonical maps $i_X:X\to X\sqcup Y$ and $i_Y:Y\to X\sqcup Y$.
So, in order to say that $Z$ is also a coproduct, I need to provide similar structure (that is, maps $j_X:X\to Z$ and $j_Y:Y\to Z$).
To do so, we just follow our nose: define $j_X$ to be the composite $X\xrightarrow{i_X}X\sqcup Y\xrightarrow fZ$, and define $j_Y$ similarly.
Now, we just need to prove that $(Z, j_X, j_Y)$ satisfies the universal property of a coproduct of $X$ and $Y$.
So, suppose we have an object $W$ with maps $k_X:X\to W$ and $k_Y:Y\to W$.
Then, we need to show that there exists a unique map $k:Z\to W$ such that the diagram
$\require{AMScd}$
\begin{CD}
X @>j_X>> Z @<j_Y<< Y \\
@VVV @V\exists!kVV @VVV \\
@>>k_X> W @<<k_Y<
\end{CD}
commutes (that is, $k_X=k\circ j_X$ and $k_Y=k\circ j_Y$).
To prove this, we need to piggyback off the fact that $X\sqcup Y$ is a coproduct, and we have an isomorphism $f:X\sqcup Y\to Z$.
Indeed, we get from the universal property of $X\sqcup Y$ a unique map $u:X\sqcup Y\to W$ such that $k_X=u\circ i_X$ and $k_Y=u\circ i_Y$, so we can construct the morphism $k:Z\to W$ as the composite $Z\xrightarrow{f^{-1}}X\sqcup Y\xrightarrow uW$.
Indeed, we see that $k\circ j_X = (u\circ f^{-1})\circ(f\circ i_X) = u\circ i_X = k_X$ and similarly for $Y$.
Conversely, if we have a map $k':Z\to W$ such that $k'\circ j_X=k_X$ and $k'\circ j_Y=k_Y$, then the map $k'\circ f:X\sqcup Y\to W$ must coincide with the universal map $u:X\sqcup Y\to W$ (by uniqueness of $u$), meaning $k'\circ f=u$ or in other words $k'=u\circ f^{-1} = k$.
This proves the uniqueness of $k$.
Let me summarise all of this work:
Proposition. Let $X\sqcup Y$ be some coproduct of $X$ and $Y$, and let $f:X\sqcup Y\to Z$ be any isomorphism. Then, $Z$ can be made into a coproduct of $X$ and $Y$.
In particular, $f$ is no longer necessarily unique as an isomorphism in the category.
Instead, the statement is different: once you give $Z$ the structure of a coproduct in the way described above, only then is $f:X\sqcup Y\to Z$ the unique isomorphism that is compatible with the coproduct structure.
(In fact, there is a correspondence between coproduct structures on $Z$ and isomorphisms $X\sqcup Y\to Z$.)
You may notice that the proof of the proposition for coproducts is very similar to the proof for the initial object provided in your question. This is because they are both instances of a more general proof for arbitrary colimits (anything isomorphic to a colimit can be made into a colimit).
This is why the proof for the fact that any object isomorphic to an initial object is initial does not require the uniqueness of the isomorphism. It just happens to be unique by the universal property of initial objects, as described in the comments!
A: You're correct: any object equipped with any isomorphism to an initial object is initial. Therefore, the given isomorphism was unique. It is not necessary to assume that the isomorphism was unique beforehand.
