Is the identity function the only order-preserving isomorphism between a linearly-ordered set and itself?

Question

It is well known that

the identity function is the only order-preserving isomorphism between a well-ordered set and itself.

My question is: what is an counterexample to this with a linearly (and thus not well-ordered) set ?

EDIT

My trivial questions has a continuation: why there is no faithful functor from $\mathbf{Lin}$ of linearly ordered sets and order preserving injective mappings to well-ordered sets and the order preserving injective mappings by the 2 examples of iso-rigidness of WOS and iso-nonrigidness of $\mathbf{Lin}$ ?

Answer 1

Re: your edit, the point is that if $\lambda$ is a well-order then there is exactly one invertible morphism from $\lambda$ to itself, but something like $\mathbb{Z}$ has many invertible self-morphisms (e.g. $x\mapsto x+1$). Now by definition a faithful functor is injective on morphisms and sends identity morphisms to identity morphisms, so there's no "room" to send $\mathbb{Z}$ to a well-ordering in a faithful way.

More generally, if $F:\mathcal{C}\rightarrow \mathcal{D}$ is faithful and every object in $\mathcal{D}$ has no non-identity invertible self-morphisms then every object in $\mathcal{C}$ must also have no non-identity invertible self-morphisms. Note that this isn't really any different from the fact that if $F:\mathcal{C}\rightarrow\mathcal{D}$ is faithful and every object in $\mathcal{D}$ has at most $17$ self-morphisms, then every object in $\mathcal{C}$ must also have at most $17$ self-morphisms, and so forth.

Answer 2

This is clearly not true. For example, $x \mapsto x+c$ is an order isomorphism for any real constant $c$, and $x \mapsto dx$ is an order isomorphism for any positive real constant $d$.