[proofplan] We compare two order isomorphisms $f$ and $g$ by forming the self-map $h = g^{-1} \circ f$ of the well-ordered set $A$. This map is an order automorphism of $(A, <_A)$. We then prove directly that every order automorphism of a well-order is the identity, using the least element of the set of points moved by the automorphism. Once $h = \operatorname{id}_A$, the equality $f = g$ follows immediately from $f = g \circ h$. [/proofplan]

[step:Construct an order automorphism of the source well-order] Because $g$ is an order isomorphism from $(A, <_A)$ onto $(B, <_B)$, it is bijective, so its inverse is a map \begin{align*} g^{-1}: B &\to A. \end{align*} Define \begin{align*} h: A &\to A \\ a &\mapsto g^{-1}(f(a)). \end{align*} Since $f$ and $g^{-1}$ are bijections, $h$ is a bijection. We verify that $h$ preserves and reflects the order $<_A$. Let $a_1, a_2 \in A$. Since $g$ is an order isomorphism and $g(h(a_i)) = f(a_i)$ for $i \in \{1,2\}$, we have \begin{align*} h(a_1) <_A h(a_2) &\iff g(h(a_1)) <_B g(h(a_2)) \\ &\iff f(a_1) <_B f(a_2) \\ &\iff a_1 <_A a_2. \end{align*} Thus $h$ is an order automorphism of $(A, <_A)$. [/step]

[step:Show that an order automorphism of a well-order fixes every element]We prove that $h = \operatorname{id}_A$. Define the subset of moved points \begin{align*} S := \{a \in A : h(a) \neq a\}. \end{align*} Suppose, toward a contradiction, that $S$ is nonempty. Since $<_A$ is a well-order, there exists an element $a_0 \in S$ such that $a_0 \leq_A s$ for every $s \in S$. Hence, for every $x \in A$ with $x <_A a_0$, we have $x \notin S$, and therefore $h(x) = x$. Because $a_0 \in S$, we have $h(a_0) \neq a_0$. By trichotomy for the total order $<_A$, either $h(a_0) <_A a_0$ or $a_0 <_A h(a_0)$. First suppose $h(a_0) <_A a_0$. Then $h(a_0) \notin S$, so $h(h(a_0)) = h(a_0)$. Since $h$ is injective, this implies $h(a_0) = a_0$, contradicting $a_0 \in S$. Now suppose $a_0 <_A h(a_0)$. Since $h$ is surjective, choose $c \in A$ such that $h(c) = a_0$. If $c <_A a_0$, then $c \notin S$, so $h(c) = c$, hence $c = a_0$, contradicting $h(a_0) \neq a_0$. Also $c \neq a_0$, since $h(c) = a_0$ and $h(a_0) \neq a_0$. Therefore trichotomy gives $a_0 <_A c$. Since $h$ preserves order, \begin{align*} h(a_0) <_A h(c) = a_0, \end{align*} contradicting $a_0 <_A h(a_0)$. Both cases are impossible, so $S$ must be empty. Therefore $h(a) = a$ for every $a \in A$, which means $h = \operatorname{id}_A$.[/step]

[guided]We want to prove that the order automorphism $h: A \to A$ cannot move any element. The well-ordering hypothesis gives a decisive way to argue: if any element is moved, there is a least moved element. Define \begin{align*} S := \{a \in A : h(a) \neq a\}. \end{align*} This is the set of elements of $A$ moved by $h$. Suppose, toward a contradiction, that $S$ is nonempty. Since $<_A$ is a well-order, every nonempty subset of $A$ has a least element, so $S$ has a least element. Let $a_0 \in S$ denote this least moved element. By the definition of $a_0$, every element strictly below $a_0$ is fixed by $h$: if $x <_A a_0$, then $x \notin S$, so $h(x) = x$. Now $a_0$ itself is moved, so $h(a_0) \neq a_0$. Since $<_A$ is a total order, exactly one of the following two possibilities holds: \begin{align*} h(a_0) <_A a_0 \end{align*} or \begin{align*} a_0 <_A h(a_0). \end{align*} Consider the first possibility, $h(a_0) <_A a_0$. Since $h(a_0)$ lies below the least moved element $a_0$, it is fixed by $h$. Hence \begin{align*} h(h(a_0)) = h(a_0). \end{align*} But $h$ is injective, because it is an order automorphism. Applying injectivity to the equality $h(h(a_0)) = h(a_0) = h(a_0)$ gives \begin{align*} h(a_0) = a_0, \end{align*} which contradicts $a_0 \in S$. Consider the second possibility, $a_0 <_A h(a_0)$. Since $h$ is surjective, there exists $c \in A$ such that \begin{align*} h(c) = a_0. \end{align*} If $c <_A a_0$, then $c$ lies below the least moved element, so $h(c) = c$. Combining this with $h(c) = a_0$ gives $c = a_0$, which is impossible because $h(a_0) \neq a_0$. Also $c \neq a_0$ directly, since $h(c) = a_0$ but $h(a_0) \neq a_0$. Therefore totality of the order forces \begin{align*} a_0 <_A c. \end{align*} Now use the order-preserving property of $h$. From $a_0 <_A c$, we obtain \begin{align*} h(a_0) <_A h(c) = a_0. \end{align*} This contradicts the current case assumption $a_0 <_A h(a_0)$. Both possible order positions of $h(a_0)$ lead to contradictions. Hence the set $S$ of moved points is empty, and therefore $h(a) = a$ for every $a \in A$. Thus $h = \operatorname{id}_A$.[/guided]

[step:Conclude that the two order isomorphisms agree] For every $a \in A$, the definition of $h$ gives $h(a) = g^{-1}(f(a))$. Applying $g$ to both sides gives \begin{align*} g(h(a)) = f(a). \end{align*} Since $h = \operatorname{id}_A$, we have $h(a) = a$ for every $a \in A$, and therefore \begin{align*} g(a) = f(a) \end{align*} for every $a \in A$. Hence $f = g$ as maps from $A$ to $B$. [/step]