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Take Schmidt decompositions of the two purifications across $A:R$. Their squared Schmidt coefficients and their $A$-side eigenspaces are fixed by the same reduced state ${\rho }_A$. The reference-side Schmidt vectors are two orthonormal families, so a unitary $U_R$ maps one family to the other, including arbitrary choices inside degenerate subspaces. Hence the unitary freedom of purification givesIf the reference supports have different dimensions, the corresponding statement uses an isometry.

Construct ensemble purificationsThey have the same reduced state exactly when ${\rho }_1={\rho }_2$. By the unitary freedom of purification, this holds exactly when $|{\Phi }_2⟩=(I\otimes U)|{\Phi }_1⟩$ for a unitary $U$. Comparing reference-basis coefficients gives the Hughston–Jozsa–Wootters theorem relationConversely, substituting this relation and using $U^{†}U=I$ immediately gives ${\rho }_2={\rho }_1$.