Unique Isomorphism between Ordinal Subset and Unique Ordinal

Theorem
Let $A \subset \operatorname{On}$ and let $A$ be a set.

Then, there is a unique mapping $f$ and a unique ordinal $x$ such that $f : x \to A$ is an order isomorphism.

Proof
Since $A \subset \operatorname{On}$, $( A, \in )$ is a strict well-ordering.

Then, the desired result follows directly from Strict Well-Ordering Isomorphic to a Unique Ordinal Under Unique Function.