Order Isomorphism is Equivalence Relation
Theorem
Order isomorphism between ordered sets is an equivalence relation.
So any given family of ordered sets can be partitioned into disjoint classes of isomorphic sets.
Proof 1
Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$ denote that $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {S_2, \preccurlyeq_2}$.
Checking in turn each of the criteria for equivalence:
Reflexivity
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Then $\struct {S, \preccurlyeq}$ is isomorphic to itself.
Thus $\cong$ is seen to be reflexive.
$\Box$
Symmetric
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1}$ be isomorphic to $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_2, \preccurlyeq_2}$ is isomorphic to $\struct {S_1, \preccurlyeq_1}$.
Thus $\cong$ is seen to be symmetric.
$\Box$
Transitive
Let $\struct {S_1, \preccurlyeq_1}$, $\struct {S_2, \preccurlyeq_2}$ and $\struct {S_3, \preccurlyeq_3}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1}$ be isomorphic to $\struct {S_2, \preccurlyeq_2}$.
Let $\struct {S_2, \preccurlyeq_2}$ be isomorphic to $\struct {S_3, \preccurlyeq_3}$.
Then $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {S_3, \preccurlyeq_3}$.
Thus $\cong$ is seen to be transitive.
$\Box$
$\cong$ has been shown to be reflexive, symmetric and transitive.
Hence the result.
$\blacksquare$
Proof 2
An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism.
The result follows directly from Relation Isomorphism is Equivalence Relation.
$\blacksquare$