[proofplan]
We prove the contrapositive by assuming that an order automorphism $f: A \to A$ moves at least one element. The moved elements then form a nonempty subset of the well-ordered set $A$, so they have a least element $c$. Every element below $c$ is fixed, and comparing $c$ with $f(c)$ gives contradictions in both possible strict-order directions, using first $f$ and then $f^{-1}$.
[/proofplan]
[step:Choose the least element moved by the automorphism]
Let $f: A \to A$ be an order isomorphism. Define the moved set
\begin{align*}
M := \{a \in A : f(a) \neq a\}.
\end{align*}
Suppose, for contradiction, that $M \neq \varnothing$. Since $(A,<)$ is well-ordered, the nonempty subset $M \subset A$ has a least element; denote it by $c \in M$.
By the definition of least element, if $a \in A$ and $a < c$, then $a \notin M$. Hence
\begin{align*}
a < c \implies f(a)=a.
\end{align*}
Since $c \in M$, we have $f(c) \neq c$.
[guided]
Let $f: A \to A$ be an order isomorphism. We want to prove that $f$ fixes every element of $A$. To use the well-ordering hypothesis, we isolate the first possible failure of being fixed.
Define
\begin{align*}
M := \{a \in A : f(a) \neq a\}.
\end{align*}
This is the subset of elements moved by $f$. Suppose, for contradiction, that $M$ is nonempty. Because $(A,<)$ is well-ordered, every nonempty subset of $A$ has a least element. Therefore $M$ has a least element; call it $c \in M$.
The meaning of this choice is important: no element strictly below $c$ is moved. Thus, for every $a \in A$,
\begin{align*}
a < c \implies a \notin M \implies f(a)=a.
\end{align*}
At the same time, because $c \in M$, we have
\begin{align*}
f(c) \neq c.
\end{align*}
So $c$ is the first element at which $f$ could fail to be the identity.
[/guided]
[/step]
[step:Rule out the possibility that $f(c)$ lies below $c$]
Assume first that $f(c) < c$. Since every element below $c$ is fixed, applied to $a=f(c)$ this gives
\begin{align*}
f(f(c)) = f(c).
\end{align*}
Because $f$ is injective, applying injectivity to the equality $f(f(c))=f(c)=f(c)$ gives
\begin{align*}
f(c)=c,
\end{align*}
contradicting $c \in M$. Therefore $f(c) < c$ is impossible.
[guided]
First consider the case in which the image of $c$ lies strictly below $c$:
\begin{align*}
f(c) < c.
\end{align*}
But every element below $c$ is fixed by the previous step. Since $f(c)$ is an element of $A$ and $f(c)<c$, we may apply that conclusion to $a=f(c)$ and obtain
\begin{align*}
f(f(c)) = f(c).
\end{align*}
Now use the injectivity of $f$, which is part of being an order isomorphism. The equality above says that two inputs, namely $f(c)$ and $c$, have the same image under $f$:
\begin{align*}
f(f(c)) = f(c).
\end{align*}
Injectivity forces the inputs to be equal, so
\begin{align*}
f(c)=c.
\end{align*}
This contradicts $c \in M$, which means precisely that $f(c)\neq c$. Hence the case $f(c)<c$ cannot occur.
[/guided]
[/step]
[step:Rule out the possibility that $c$ lies below $f(c)$]
Assume next that $c < f(c)$. Since $f^{-1}: A \to A$ is also an order isomorphism, applying $f^{-1}$ to the inequality $c < f(c)$ gives
\begin{align*}
f^{-1}(c) < c.
\end{align*}
The element $f^{-1}(c)$ is below $c$, so it is fixed by $f$:
\begin{align*}
f(f^{-1}(c)) = f^{-1}(c).
\end{align*}
But $f(f^{-1}(c))=c$, so $c=f^{-1}(c)$. Applying $f$ to both sides gives $f(c)=c$, contradicting $c \in M$. Therefore $c < f(c)$ is impossible.
[guided]
Now consider the opposite strict inequality:
\begin{align*}
c < f(c).
\end{align*}
Because $f: A \to A$ is an order isomorphism, its inverse map $f^{-1}: A \to A$ is also an order isomorphism. Applying the order-preserving map $f^{-1}$ to both sides of $c < f(c)$ gives
\begin{align*}
f^{-1}(c) < f^{-1}(f(c)) = c.
\end{align*}
Thus $f^{-1}(c)$ is an element strictly below $c$.
By the choice of $c$ as the least moved element, every element below $c$ is fixed by $f$. Therefore
\begin{align*}
f(f^{-1}(c)) = f^{-1}(c).
\end{align*}
But by the definition of inverse function,
\begin{align*}
f(f^{-1}(c)) = c.
\end{align*}
Combining these equalities gives
\begin{align*}
c = f^{-1}(c).
\end{align*}
Applying $f$ to both sides yields
\begin{align*}
f(c)=c,
\end{align*}
again contradicting $c \in M$. Hence the case $c<f(c)$ cannot occur.
[/guided]
[/step]
[step:Use totality of the order to conclude that no element is moved]
Since $(A,<)$ is a well-order, it is in particular a strict total order. For the two elements $c$ and $f(c)$, exactly one of the following alternatives holds:
\begin{align*}
f(c)<c,\qquad f(c)=c,\qquad c<f(c).
\end{align*}
The first and third alternatives have been ruled out, while the middle alternative contradicts $c \in M$. This contradiction shows that $M=\varnothing$. Therefore $f(a)=a$ for every $a \in A$, so $f=\operatorname{id}_A$.
[guided]
The proof has now eliminated both possible strict comparisons between $c$ and $f(c)$. To finish, we use the order-theoretic fact that a well-order is a strict total order: for any two elements of $A$, exactly one of the three relations holds. Applying this to the two elements $c$ and $f(c)$, we have
\begin{align*}
f(c)<c,\qquad f(c)=c,\qquad c<f(c).
\end{align*}
The case $f(c)<c$ was ruled out by injectivity of $f$ and minimality of $c$. The case $c<f(c)$ was ruled out by applying the inverse order isomorphism $f^{-1}$ and again using minimality of $c$. The remaining case is
\begin{align*}
f(c)=c,
\end{align*}
but this contradicts $c \in M$, since $M$ was defined by
\begin{align*}
M := \{a \in A : f(a) \neq a\}.
\end{align*}
Thus the assumption $M \neq \varnothing$ is impossible, so $M=\varnothing$. Therefore no element of $A$ is moved by $f$: for every $a \in A$,
\begin{align*}
f(a)=a.
\end{align*}
Hence $f=\operatorname{id}_A$, which proves that the only order isomorphism from $A$ to itself is the identity map.
[/guided]
[/step]
