Unique Isomorphism between Ordinal Subset and Unique Ordinal

Theorem
Let $\operatorname{On}$ be the class of ordinals.

Let $S \subset \operatorname{On}$ where $S$ is a set.

Then there exists a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi : x \to S$ is an order isomorphism.

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

The result follows directly from Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping.