Let $A=[0,\omega_0]$ and $K=[0,1]$ be closed intervals of the ordinals and the real numbers with natural orderings. Let $L=A\times K$ be equipped with the lexicographic order where the $A$ coordinate is more important. Then both $L$ and $K$ are complete and the injective homomorphisms are easy to construct. To see that $L\not\cong K$ as ordered sets observe that $K$ is connected in its order topology while $L$ is not.

Let $A=[0,\omega_0]$ and $K=[0,1]$ be closed intervals of the ordinals and the real numbers with natural orderings. Let $L=A\times K$ be equipped with the lexicographic order where the $A$ coordinate is more important. Then both $L$ and $K$ are complete and the injective homomorphisms are easy to construct. To see that $L\not\cong K$ as ordered sets observe that $K$ is connected in its order topology while $L$ is not.

Edit: Please unaccept this answer so that I can delete it. The correct answer is provided by Eric Wofsey. The $K$ admits no complete embedding into $L$ as such an embedding would have to preserve the least and the largest elements (inf and sup of the empty set).