For countable models, the answer is yes, since all nonstandard such models have the same order type. (For the standard model $N=mathbb{N}$, assume there is an $omega$-standard model of ZFC, and in general we must assume Con(ZFC).) Namely, if $Nmodelstext{PA}$ is nonstandard, then its order type is $omega+mathbb{Z}cdotmathbb{Q}$, and this is also the order type appearing as the $omega$ of any $omega$-nonstandard countable model of ZFC.

It seems to me that this argument would also work for saturated models in larger cardinalities, but I’m unsure of the general case.

A related question would be: which models of PA arise as the full standard structure of arithmetic $langlemathbb{N},+,cdot,0,1,<rangle^M$ of a model $Mmodelstext{ZFC}$? In other words, we ask for the full structure, rather than merely the order type as you have. These are known as the ZFC-standard models of PA, and in the countable case they are exactly the standard model of PA (provided there is an $omega$-model of ZFC), together with the computably saturated models of the arithmetic consequences of ZFC, a theory denoted $text{Th}(mathbb{N})^{text{ZFC}}$. For a discussion of this and applications, see proposition 3 of Satisfaction is not absolute, as well as a 2009 report of Ali Enayat on the topic for the Mittlag-Lefflar institute.

Meanwhile, I was able to get the following, which works even for

uncountable models. (**Update:** I made this an equivalence.)

**Theorem.** The following are equivalent for any $Nmodelstext{PA}$.

- For any finitely many axioms $varphi_0,ldots,varphi_n$ of ZFC, the model $N$ thinks they are consistent.
- As a model of arithmetic, $N$ is isomorphic to an initial segment

of the standard model $mathbb{N}^M$ of some $Mmodelstext{ZFC}$.
- $N$ satisfies all the $Pi_1^0$ consequences of ZFC.

Proof. ($2to 3$) If $N$ is an initial segment of $mathbb{N}^M$ for some $Mmodelstext{ZFC}$, then $N$ must satisfy all the $Pi_1^0$ arithmetic statements true in $M$.

($3to 1$) Those consistency statements are amongst the $Pi_1^0$ consequences of ZFC, which proves the consistency of any particular of its finite fragments.

($1to 2$) Suppose that $N$ is a model of PA in

which every standard finite fragment of standard ZFC is consistent. (This

follows from $Nmodelstext{Con}(text{ZFC})$, but it is a strictly

weaker hypothesis.) In particular, this hypothesis implies that ZFC is consistent, and so we may assume without loss that $N$ is nonstandard. Consider $N$’s version of the theory ZFC. We

may enumerate what it thinks are the axioms, which of course

includes many nonstandard axioms. Since we have assumed that $N$

thinks all the standard-finite initial segments of this

enumeration are consistent, by overspill there is some nonstandard

finite theory $t$ in $N$ which $N$ thinks is consistent, and which

includes every standard axiom of ZFC.

Inside $N$, we may now build the usual Henkin completion $T$ of

$t$. That is, we add Henkin constants and add every sentence

$sigma$ or its negation $negsigma$ in the new language, in such

a way that $N$ thinks $T$ extends $t$ and remains consistent,

while having the Henkin property. The theory $T$ is a definable

class in $N$, and $N$ is also able to define the corresponding

Henkin model $M$.

By meta-theoretic induction, we can show that $Mmodelssigma$

just in case $sigmain T$, for every standard finite sentence

$sigma$ in the expanded language. In particular, it follows that

$M$ is a model of ZFC.

Finally, I claim that $N$ is isomorphic to an initial segment of

the $mathbb{N}^M$, and indeed, $N$ is able to construct this

isomorphism. For every $ain N$, the model $N$ is able to

construct the term $overbrace{1+cdots+1}^a$, and it has added a

Henkin constant $hat a$ and the assertion $hat

a=overbrace{1+cdots+1}^a$ to the theory $T$. Furthermore, $N$

believes that it is able to prove (by induction) that $x<hat a$

implies $x=hat b$ for some $b<a$. It also must believe things

like $hat{(a+b)}=hat a+hat b$ and $hat{(acdot b)}=hat

acdothat b$, since these kind of manipulations of terms are

provable in PA. Thus, the map $$amapsto (hat a)^M$$ is an

isomorphism of $langle N,+,cdot,0,1,<rangle$ to an initial

segment of $mathbb{N}^M$, as desired. QED

**Corollary.** If ZFC is consistent, then every model of true arithmetic, and indeed every model of $text{PA}+text{Con}(text{ZFC})$, arises as an initial segment of the standard model $mathbb{N}^M$ of some model $Mmodelstext{ZFC}$ of set theory.

One can replace ZFC in this argument with other set theories, such as KP or ZFC + large cardinals, and get an equivalence for other theories.

Basically, in the implication ($1to 2$) of the theorem, the model $N$ thinks it

has constructed an $omega$-nonstandard model of the theory $t$,

and it naturally believes that what it thinks is the standard

model, itself, is an initial segment of that model.