Order-Preserving Identity Mapping between Ordered Structures not necessarily Isomorphism

Theorem
Let $A$ be a set.

Let $\RR$ and $\SS$ be orderings on $A$ such that:
 * $\forall a, b \in A: a \mathrel \RR b \implies a \mathrel \SS b$

Let $I_A$ denote the identity mapping on $A$.

Then it is not necessarily the case that $I_A$ is an order isomorphism from the ordered structures $\struct {A, \RR}$ and $\struct {A, \SS}$.

Proof
Proof by Counterexample

Let $\RR: \N \to \N$ be the ordering on the natural numbers $\N$ defined as:


 * $\forall a, b \in \N: a \mathrel \RR b \iff a \le b \text { and } \map P a = \map P b$

where $\map P x$ is the parity of $x$.

That is:
 * $0 \mathrel \RR 2 \mathrel \RR 4 \mathrel \RR \cdots$

and:
 * $1 \mathrel \RR 3 \mathrel \RR 5 \mathrel \RR \cdots$

and so on, but (for example):
 * $\lnot \paren {0 \mathrel \RR 1}$

Let $\SS: \N \to \N$ be the ordering on the natural numbers $\N$ defined as:
 * $\forall a, b \in \N: a \mathrel \SS b \iff a \le b$

We have that:

But note that:

while it is not the case that $0 \mathrel \RR 1$.

Hence while it is true that:
 * $\forall a, b \in \N: a \mathrel \RR b \implies a \mathrel \SS b$

it is not the case that:
 * $\forall a, b \in \N: a \mathrel \SS b \implies a \mathrel \RR b$

and so the identity mapping on $\NN$ is not an order isomorphism.