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I am new to set theory and I have come across this exercise. I think the proof I have come up with is too simple to be right. would greatly appreciate some tips.

If $I$ is an initial set, then for $\forall B$ in a category C, $\exists !$ an arrow $I \to B$. Since $I'$ is a set in $C$, $\exists !$ arrow to A such that $I \to I'$. The same theory can be applied to prove that there is an arrow $B$ such that $I' \to I$. Therefore, there is an isomorphism between $I$ and $I'$.

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You're halfway through, because you didn't show these arrows going in opposite directions are actually inverses. Assume $I$ and $I'$ are initial. Then since $I$ is initial, there is a unique arrow $f:I \to I'$. Since $I'$ is initial, there is a unique arrow $g:I' \to I$. Now, both $g \circ f$ and $1_I$ are arrows $I \to I$, but since $I$ is initial, there can be only one: we conclude that $g\circ f = 1_I$. Similarly, $f \circ g = 1_{I'}$. Now you can say that $I \cong I'$.

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