Unique Isomorphism between Equivalent Finite Totally Ordered Sets

Theorem
Let $$S$$ and $$T$$ be finite sets such that $$\left|{S}\right| = \left|{T}\right|$$.

Let $$\left({S; \preceq}\right)$$ and $$\left({T; \preccurlyeq}\right)$$ be ordered structures on which both $$\preceq$$ and $$\preccurlyeq$$ are total orderings.

Then there is exactly one isomorphism from $$\left({S; \preceq}\right)$$ to $$\left({T; \preccurlyeq}\right)$$.

Proof
It is sufficient to consider the case where $$\left({T; \preccurlyeq}\right)$$ is $$\left({\mathbb{N}_n; \le}\right)$$ for some $$n \in \mathbb{N}$$.

Let $$A$$ be the set of all $$n \in \mathbb{N}$$ such that if:


 * 1) $$S$$ is any set such that $$\left|{S}\right| = n$$, and
 * 2) $$\preceq$$ is any total ordering on $$S$$

then there is exactly one isomorphism from from $$\left({S; \preceq}\right)$$ to $$\left({\mathbb{N}_n; \le}\right)$$.


 * $$\varnothing \in A$$ from Null Mapping and Same Cardinality Bijective Injective Surjective.


 * Now let $$n \in A$$.

Let $$\left({S; \preceq}\right)$$ be a totally ordered set with $$n + 1$$ elements.

By Finite Subset of Totally Ordered Set, $$S$$ has a maximal element $$b$$.

Then $$\left|{S - \left\{{b}\right\}}\right| = n$$ by Cardinality Less One.

The total ordering on $$S - \left\{{b}\right\}$$ is the one induced from that on $$S$$.

So as $$n \in A$$, there exists a unique isomorphism $$f: S - \left\{{b}\right\} \to \left({\mathbb{N}_n; \le}\right)$$.

Let us define the mapping $$g: S \to \mathbb{N}_{n+1}$$ as follows:

$$\forall x \in S: g \left({x}\right) = \begin{cases} f \left({x}\right): & x \in S - \left\{{b}\right\} \\ n: & x = b \end{cases} $$

This is the desired isomorphism from $$\left({S; \preceq}\right)$$ to $$\left({\mathbb{N}_{n+1}; \le}\right)$$.

Now let $$h: \left({S; \preceq}\right)$$ to $$\left({\mathbb{N}_{n+1}; \le}\right)$$ be an isomorphism.

Then $$h \left({b}\right) = n$$ (it has to be).

So the restriction of $$h$$ to $$S - \left\{{b}\right\}$$ is an isomorphism from $$S - \left\{{b}\right\}$$ to $$\left({\mathbb{N}_n; \le}\right)$$, and hence $$h = f$$ (as $$n \in A$$, any such isomorphism is unique).

Thus:

$$\forall x \in S - \left\{{b}\right\}: h \left({x}\right) = f \left({x}\right) = g \left({x}\right)$$

and:

$$h \left({b}\right) = n = g \left({b}\right)$$

thus $$h = g$$.

Therefore $$n + 1 \in A$$.


 * The result follows from the Principle of Finite Induction.