is $\omega$ in every model of ZFC always isomorphic to the standard natural numbers

Question

ZFC can have some very odd models one of them being countable. Is it true that in any model of ZFC $\omega$ with the empty set as zero the successor function $S(x):= x\cup\{x\}$ and the addition and moltiplication functions defined by recursion is isomorphic to the "real" natural numbers. I know that Second order arithmetic with quantifiers over all the subsets of it's domain is categorical but from my understanding in some ZFC models not all of the subsets of $\omega$ are sets

Answer 1

No (assuming ZFC is consistent). This is essentially immediate from the compactness theorem. For instance, add uncountably many constant symbols $c_i$ to the language of ZFC, and add axioms saying that $c_i\neq c_j$ for each $i\neq j$ and $c_i\in\omega$ for each $i$. Every finite subset of the resulting theory is consistent, since it will only involve finitely many of the $c_i$ and so we can send those $c_i$ to distinct natural numbers in some model of ZFC and the rest to $0$. So by compactness, this theory has a model $M$. Then $M$ is a model of ZFC such that $\omega^M$ has (externally) uncountably many elements.