Order Automorphism on Well-Ordered Class is Identity Mapping

Theorem
Let $\struct {A, \preccurlyeq}$ be a well-ordered class.

Let $\phi$ be an order isomorphism on $\struct {A, \preccurlyeq}$.

Then $\phi$ is the identity mapping:
 * $\forall a \in A: \map \phi a = a$

Proof
Let $\phi$ be an order isomorphism.

Then from Inverse of Order Isomorphism is Order Isomorphism, the inverse mapping $\phi^{-1}$ is also an order isomorphism.

From Order Automorphism on Well-Ordered Class is Forward Moving:


 * $\forall a \in A: a \preccurlyeq \map \phi a$

and:
 * $\forall a \in A: a \preccurlyeq \map {\phi^{-1} } a$

from which:
 * $\forall a \in A: \map \phi a \preccurlyeq \map \phi {\map {\phi^{-1} } a} = a$

Hence:
 * $\forall a \in A: \map \phi a = a$