Is the identity function the only order-preserving isomorphism between a well-ordered set and itself? I have a question.
Let $(x,\leq)$ be a well ordered set. let $f:x \to x$ be an isomorphism such that $a \leq b$ implies $f(a) \leq f(b)$ (meaning f is perserving the order). Is $f$ the identity function? Is $f(a)=a$ for all $a \in x$? Or are there more order-preserving isomorphisms between $x$ and itself.
 A: Let $x$ be well-ordered and  $f\colon x\to x$ be a bijection with $f(a)\le f(b)$ whenever $a\le b$.
Assume $f$ is not the identity. The set of $y\in x$ with $f(y)\ne y$ has a minimal element $a$. 
If $f(a)< a$ then $f(f(a))=f(a)$ by minimality of $a$, but that contradicts injectivity of $f$.
If $f(a)>a$ conclude from surjectivity of $f$ that $f(b)=a$ for some $b\in x$.
Clearly, as $b<a$ would lead to $a=f(b)=b<a$, we conclude $b>a$. But then $f(b)\ge f(a)>a$, contradiction.
A: Hint: Suppose that $f:X\to X$ is an order-isomorphism where $X$ is some well-ordered set, and suppose by way of contradiction that there is some least $x\in X$ such that $f(x)\ne x.$ It follows that $f\bigl(f(x)\bigr)\ne f(x)$ (why?), so that $x<f(x)$ (why?). Can you take it from here?
A: Just thought of an answer as well. Thought I'd add this for the other readers who are interested.
This is a proof by induction. Since $X$ is well ordered, it has a minimal element. assume $a_0 \in X$ is the minimal element. Because f is surjective, there is a $b_0 \in x$ such that $f(b_0)=a_0$.
Let us assume that $b_0 \neq a_0$ which would imply that $f$ is not the identity function.
Now, since $a_0$ is minimal, and $b_0 \neq a_0$, it follows that $b_0 > a_0$, and since $f$ is order perserving, we get $f(b_0) > f(a_0)$ but $f(b_0)=a_0$! so overall we get $a_0>f(a_0)$ and so $a_0$ is not the minimal element, contradiction!
Now do the same process for $X-\{a_0\}$ (which is also well ordered) until we finally reach an empty set. Deduce that $f(a)=a$.
A: Let f : X → X be an isomorphism and, towards a contradiction, suppose for some v ∈ X, f (v) $\neq$ v.
So the set
W = {v ∈ X : f (v) $\neq$ v} is nonempty. Let x be the ≺-least member of W.
Put y = f (x).
Then either y ≺ x or x ≺ y.
If
y ≺ x, then f (y) = y as x was ≺- least non-fixed point of f.
But since f preserves ≺ and y ≺ x, y = f(y) ≺ f(x) = y which is impossible.
Next, suppose x ≺ y. Since f is surjective, there is some w ∈ X such that f (w) = x.
Clearly, w $\nleq$ x so x ≺ w. But then y = f (x) ≺ f (w) = x which contradicts x ≺ y.
