Do any two ill-founded models of set theory with order isomorphic ordinals have isomorphic copies of L?

Let $M$ and $N$ be ill-founded models of set theory with $\textbf{ORD}^M$ order isomorphic to $\textbf{ORD}^N$. Does it follow that $L^M$ is isomorphic to $L^N$? If not, does isomorphism follow from a strengthening of the hypothesis? Say, $(\textbf{ORD}^M, +, \times) \cong (\textbf{ORD}^N, +, \times)$?

More generally, does transfinite recursion between models of set theory with isomorphic ordinals always result in isomorphic class being defined?

Solution 1:

The answer to the initial question is negative: very little information is captured by the ordertype of the ordinals alone. One way to see this is to note that any two countable models of $\mathsf{ZFC}$ with nonstandard $\omega$ have order-isomorphic ordinals (this is a fun exercise: to start, show that their $\omega$s each have ordertype $\mathbb{N}+\mathbb{Z}\cdot\mathbb{Q}$).

This does not at all generalize to the case where we include addition and multiplication of ordinals, however.