Given a linear order $\mathcal{S}$, let $\mathbb{A}_\mathcal{S}$ be the class of all ordertypes which do not embed $\mathcal{S}$ (= do not have a suborder isomorphic to $\mathcal{S}$). Say that a linear order $\mathcal{S}$ is deep iff there is a computable functional $\Phi$ with the property that, whenever $\mathcal{M}$ is a linear order with domain $\mathbb{N}$ and $\mathcal{M}\in\mathbb{A}_\mathcal{S}$, we have $\Phi^\mathcal{M}\in\mathbb{A}_\mathcal{S}\setminus\mathbb{A}_\mathcal{M}$ but $\mathcal{M}\in\mathbb{A}_{\Phi^\mathcal{M}}$. (Note that $\Phi$ does not have to preserve isomorphism. Also, if $\mathcal{S}$ is deep then $\mathbb{A}_\mathcal{S}$ trivially has no maximal elements, but I see no reason for the converse to hold.)
For example, $\omega, \omega^*$, and $\eta$ (= the ordertypes of the naturals, negative naturals, and rationals respectively) are each deep via the single functional $\mathcal{M}\mapsto\mathcal{M}+\mathcal{M}$. On the other hand, no finite linear order is deep for trivial reasons.
Question 1: Is $\zeta$ (= the ordertype of the integers) deep?
Certainly the same functional won't work here: $\omega+\omega^*\in\mathbb{A}_\zeta$, but $\omega+\omega^*+\omega+\omega^*=\omega+\zeta+\omega^*$.
More generally, I'm interested in which countable linear orders are deep; $\zeta$ just seems like the first interesting case to consider.
I'm also interested in the following (even-more-)effective version of depth. Say that $\mathcal{S}$ is effectively deep iff there are computable functionals $\Phi,\Psi,\Theta,\Xi$ such that for every linear order $\mathcal{M}$ whatsoever we have
$\Phi^\mathcal{M}$ is a linear order,
$\Psi^\mathcal{M}$ is an embedding of $\mathcal{M}$ into $\Phi^\mathcal{M}$,
if $f$ is an embedding of $\mathcal{S}$ into $\Psi^\mathcal{M}$, then $\Theta^{f,\mathcal{M}}$ is an embedding of $\mathcal{S}$ into $\mathcal{M}$, and
if $g$ is an embedding of $\Phi^\mathcal{M}$ into $\mathcal{M}$, then $\Xi^{g,\mathcal{M}}$ is an embedding of $\mathcal{S}$ into $\mathcal{M}$.
(In the last two bulletpoints, if $f/g$ is not of the appropriate type we don't require $\Theta^{f,\mathcal{M}}/\Xi^{g,\mathcal{M}}$ to even be fully defined.)
Question 2: Is $\zeta$ effectively deep?
(Note that $\omega,\omega^*,$ and $\eta$ are still effectively deep.)