What is the cardinality of the set of equivalence classes of countable order types under bi-embeddability? Two ordered sets have the same order type if there exists an order isomorphism between them.  Now the set $X$ of all order types of countable totally ordered sets has cardinality $2^{\aleph_0}$; see here.  But I'd like to find the cardinality of a smaller set.
Let us define an equivalence relation $\sim$ on $X$ as follows.  If $a$ is the order type of $A$ and $b$ is the order type of $B$, then we say that $a\sim b$ if there exists an order isomorphism between $A$ and a subset of $B$, and an order isomorphism between $B$ and a subset of $A$.  My question is, what is the cardinality of the set of equivalence classes of elements of $X$ under this equivalence relation?
Is it still $2^{\aleph_0}$, or is it smaller than that?
 A: The cardinality is $\aleph_1$.
As noted by Andrés Caicedo, the nonscattered countable linear orders form a single $\sim$-class, so it suffices to explain why there are only $\aleph_1$-many $\sim$-classes of scattered types. 
Rich Laver proved that the embeddability quasiorder of (countable) nonscattered types is a well-quasiorder. Call this quasiorder $W$. Let $P=W/\sim$ be the quotient poset. The poset $P$ has no infinite antichains or infinite properly descending chains. The task is to determine the size of $P$.
Recursively partition $P$ into levels: $L_0$ is the antichain of minimal elements of $P$, and for each $\alpha>0$, define $L_{\alpha}$ to be the antichain of minimal elements of $P-\bigcup_{\beta<\alpha} L_{\beta}$. Let $\gamma$ be the least ordinal that does not index any level. Then $|P| = |\bigcup_{\alpha<\gamma} L_{\alpha}| = |\gamma|$, since $\gamma$ is infinite and the levels are finite and partition $P$.
The question asks for the number of $\sim$-classes, which is the cardinality of $P$. We have just shown that this is $|\gamma|$. By Eric Wofsey's comment, $|\gamma|\geq \aleph_1$. 
To obtain a contradiction, assume that $|\gamma|>\aleph_1$. Choose an ordinal $\delta$ so that $\aleph_1<\delta<\gamma$. Level $L_{\delta}$ is not empty, since $\delta<\gamma$, so pick a countable linear order $\Lambda_{\delta}$ so that $(\Lambda_{\delta}/\sim)\in L_{\delta}$. Every level $L_{\varepsilon}$, for $\varepsilon <\delta$, contains some $\Lambda_{\varepsilon}/\sim$, where $\Lambda_{\varepsilon}$ is a countable linear order embeddable in $\Lambda_{\delta}$. (If there exists some $\varepsilon < \delta$ whose level $L_{\varepsilon}$ does not contain the $\sim$-class of any poset embeddable in $\Lambda_{\delta}$, then the least-indexed level with this property would have to contain $\Lambda_{\delta}/\sim$, contradicting either $\varepsilon <\delta$ or the disjointness of the levels.)
But now we have a countable linear order $\Lambda_{\delta}$ which has $|\delta|$-many order types of pairwise $\sim$-inequivalent, countable linear orders that are embeddable into it, namely the $\Lambda_{\varepsilon}$'s. And remember that $\delta$ is uncountable. We have now obtained a contradiction to 
Fraisse's Second Conjecture (also proved by Laver), which states that a countable scattered linear order has only countably many $\sim$-types of suborders.   \\\\\
You will find the statements of all four of Fraisse's Conjectures on page 178 of Rosenstein's book Linear Orderings, Academic Press, 1982. The proof (due to Laver) of Fraisse's Second Conjecture is Theorem 10.52 of this book.
