= Solution
Represent any universally quantum device, including its auxiliaries and arbitrary final readout, by a <positive operator-valued measure> $\{E_i\}$ on the input. Perfect identification would require
$$
E_i\geq0,\qquad \sum_iE_i=I,\qquad
\langle\phi_j|E_i|\phi_j\rangle=\delta_{ij}.
$$
Since a positive operator has a positive square root, zero expectation implies $E_i|\phi_j\rangle=0$ for $j\ne i$. Also $I-E_i\geq0$ and its expectation in $\phi_i$ is zero, so $E_i|\phi_i\rangle=|\phi_i\rangle$. Consequently
$$
\langle\phi_i|\phi_j\rangle
=\langle\phi_i|E_i|\phi_j\rangle=0
\qquad(i\ne j).
$$
This proves that <perfect discrimination of pure states requires orthogonality>. Any nonzero overlap contradicts perfect identification, independently of whether the device disturbs its input.
Equivalently, a <unitary operator> coupling the input to a readout apparatus must preserve inner products. Distinct perfectly readable records have zero inner product, so their corresponding input vectors must already be orthogonal. \b[Nonorthogonal candidate states cannot be identified with certainty in one use.] Probabilistic <unambiguous quantum state discrimination> can have an inconclusive outcome, which is excluded by the guaranteed-identification requirement.
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