The normalization is unique up to unique isomorphism

First suppose $X$ is irreducible. Each ${\nu }_i$ is a dominant birational morphism from the normal variety $X_i$ [F1]. Apply the universal property [F2] to the normalization ${\nu }_1$ and the morphism ${\nu }_2\text{\hspace{0.17em}⁣}:X_2\to X$: there is a unique morphism $g\text{\hspace{0.17em}⁣}:X_2\to X_1$ over $X$, i.e. with ${\nu }_1\circ g={\nu }_2$. Symmetrically there is a unique morphism $h\text{\hspace{0.17em}⁣}:X_1\to X_2$ with ${\nu }_2\circ h={\nu }_1$.

The composite $h\circ g\text{\hspace{0.17em}⁣}:X_2\to X_2$ satisfies ${\nu }_2\circ (h\circ g)={\nu }_1\circ g={\nu }_2$. Apply [F2] to the normalization ${\nu }_2:X_2\to X$ and the morphism ${\nu }_2:X_2\to X$: both $h\circ g$ and ${\mathrm{i}\mathrm{d}}_{X_2}$ lift that morphism, so uniqueness gives $h\circ g={\mathrm{i}\mathrm{d}}_{X_2}$. Symmetrically $g\circ h={\mathrm{i}\mathrm{d}}_{X_1}$, so $g$ is an isomorphism with inverse $h$.

Any isomorphism $u\text{\hspace{0.17em}⁣}:X_1\to X_2$ over $X$ satisfies ${\nu }_2\circ u={\nu }_1$, so $u$ is a morphism over $X$ lifting the identity of $X$ between the two normalizations; by the uniqueness clause of [F2] applied to the normalization ${\nu }_2$ and the dominant birational morphism ${\nu }_1:X_1\to X$, such a $u$ equals $h$ constructed above. Thus the isomorphism is unique in the irreducible case. For reducible $X$, [F1] describes each normalization as the disjoint union over its finitely many irreducible components. Apply the irreducible result to each component and take the disjoint union. Any map over $X$ sends a source component into the target normalization component with the same dense image in $X$, since the target components are disjoint; uniqueness therefore holds componentwise. If $X$ is empty both normalizations are empty.