 Solution

In a countable first-order model, the finitely many named constants split the order into the open intervalstogether with the $n$ named points. Each interval is a countable dense linear order without endpoints in its induced order. It is nonempty and infinite by density and the absence of endpoints.

Use Cantor's classification of countable dense linear orders without endpoints: each such interval is isomorphic to the rational order. It follows by a back-and-forth method extending finite increasing partial bijections alternately along enumerations of the two orders. Given two first-order models of ${\mathrm{DLO}}_n$, choose isomorphisms between their corresponding intervals and send each named point to its counterpart. Their union is an increasing bijection preserving all constants, hence an isomorphism in $L_n$.

There are countable first-order models, for example any finite increasing list in the rational order. ThereforeFor $n=0$, the same argument is just the classification theorem for the whole order.