In my axiomatic set theory notes, there appears that, if $A$ and $B$ are well-ordered isomorphic sets, then there exists one isomorphism between them. However, as a side note, it is stated that this fact apparently distinguishes only-linearly orderded sets from well-ordered ones.
In the case where $A=B$, that is, considering the special case where we are studying the existence of automorphisms on $A$ that are different from the identity, $\text{id}_{A}$, one can easily find that, in $\mathbb{R}$ with its usual order there are plenty of counterexamples that prove this claim; for instance, considering different translations of the points of $\mathbb{R}$
However, I have not yet encountered a single counterexample that proves this in the case where $A\not=B$, that is, a couple of distinct linearly ordered sets that are isomorphic via several dstinct automorphisms.
I would guess that this counterexample should involve some subsets of $\mathbb{R}$, because the majority of sets we have encountered in this course, like the ordinal numbers, are all well ordered sets with their natural order relation.
Thanks in advance for your time.
Fix an order-isomorphism $(0,1) \to \mathbb{R}$ and compose it with any order-isomorphism $\mathbb{R} \to \mathbb{R}$. Each order-isomorphism $\mathbb{R} \to \mathbb{R}$ then gives rise to a distinct order-isomorphism $(0,1) \to \mathbb{R}$.