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Suppose $F=GJ=G^{\prime }{J}^{\prime }$ are two final-discrete-fibration factorizations. Apply the orthogonality of final functors and discrete fibrations to the square with left side $J$, top $J^{\prime }$, bottom $G$ and right side $G^{\prime }$. It gives a unique $S:\mathcal{E}\to {\mathcal{E}}^{\prime }$ satisfying $SJ=J^{\prime }$ and $G^{\prime }S=G$. Reversing the roles gives $T:{\mathcal{E}}^{\prime }\to \mathcal{E}$.

Both $TS$ and $1_{\mathcal{E}}$ fill the same square for $J,G$, so uniqueness gives $TS=1_{\mathcal{E}}$; similarly $ST=1_{{\mathcal{E}}^{\prime }}$. This proves the uniqueness of a final-discrete-fibration factorization: the connecting functor is a canonical isomorphism, not just a noncanonical equivalence.