TITLE: Is every order type of a PA model the \omega of some ZFC model?
QUESTION [11 upvotes]: Let $N$ be a model of first-order Peano arithmetic, and let $\sigma$ be its order-type. Does it follow that there is a (non-transitive, expect when $M$ is the standard model) $ZFC$-model $M$ such that $\omega^{M}=\sigma$? What if $PA$ is strengthened to true arithmetic (i.e. the theory of $(\mathbb{N},+,\cdot)$ or weakened to a subtheory such as $I\Sigma_{1}$, $I\Delta_{0}+EXP$, $I\Delta_{0}$ or $IOpen$? Or when we replace $ZFC$ with something weaker like $KP$?

REPLY [10 votes]: Joel Hamkins has already answered the question in the positive for countable models of PA; indeed PA can be relaxed to IOpen in his argument.
In the uncountable case, as far as I know, the problem is wide open. 
My reason: a closely related question is still (wide) open, namely, is it possible to have distinct consistent completions $T_1$ and $T_2$ of PA such that the collection of order-types of uncountable models of $T_1$ differs from the collection of order-types of uncountable models of $T_2$ ?
This open question was first brought to world attention by Harvey Friedman in his 1975-paper "One hundred and two problems in mathematical logic" (in Journal of Symbolic Logic).
For the state of the art (in relation to order-types of models of PA) see Chapter 11 of the Kossak-Schmerl graduate text on on models of arithmetic.  Friedman's problem is mentioned there both in Chapter 11, and also in Chapter 12 (List of 20 open questions).