= Solution
A <well-order> is a linearly ordered set in which every nonempty subset has a least element. First note that a <well-order> cannot be isomorphic to a proper <initial segment> of itself: the least element moved by such an isomorphism would have exactly the same predecessor set on each side, forcing it to be fixed. The same least-disagreement argument makes an isomorphism between <initial segments> unique.
Match $a\in A$ to $b\in B$ exactly when their strict predecessor sets are order-isomorphic. The preceding observation makes each match unique. Matches preserve order and fill <initial segments> on both sides, since restricting an isomorphism matches every earlier element. If both sides had an unmatched element, take the least unmatched $a$ and $b$. Their predecessor sets would be exactly the already matched segments, so they too would match, a contradiction. Therefore the matching exhausts at least one <well-order> and gives an isomorphism of that whole set onto an <initial segment> of the other. Uniqueness follows from the same predecessor-set argument. The whole set is allowed as an <initial segment>; this includes equal <order types>.
For <ordinal> division, let $\beta>0$ and choose
$$
\gamma=\sup\{\eta:\beta\eta\leq\alpha\}.
$$
This supremum is a set <ordinal> because $\beta\eta\geq\eta$ bounds the displayed <ordinals> by $\alpha$. Strict increase and <continuity> of multiplication in the right argument imply $\beta\gamma\leq\alpha<\beta(\gamma+1)$. Take the <order type> $\delta$ of the remaining tail of $\alpha$ after its <initial segment> of type $\beta\gamma$. Then $\alpha=\beta\gamma+\delta$; if $\delta\geq\beta$ it would contradict the strict upper inequality. Thus
$$
\boxed{\alpha=\beta\gamma+\delta,\qquad\delta<\beta.}
$$
The half-open intervals $[\beta\eta,\beta(\eta+1))$ are disjoint, so $\gamma$ is unique, and uniqueness of the tail gives uniqueness of $\delta$. This is left-divisor <ordinal> division: the order of multiplication cannot be interchanged.
Apply it with $\beta=\omega$. The remainder is finite. A nonzero remainder makes $\omega\gamma+\delta$ a successor, while a nonzero $\omega\gamma$ is a limit: for successor $\gamma$ it ends in a full block of type $\omega$, and for limit $\gamma$ it is the supremum of the earlier blocks. Consequently
$$
\boxed{\lambda\ne0\text{ is a limit ordinal}\quad\Longleftrightarrow\quad\lambda=\omega\gamma\text{ for some }\gamma>0.}
$$
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