= Solution
The root $\omega$ must be a \b[primitive $n$th root of unity], for example $e^{2\pi i/n}$. The printed statement merely says $n$th root; that is insufficient. With $n=2$ and $\omega=1$, the states with $r=0$ and $r=1$ coincide. More generally, a root of order $d<n$ repeats the $r$ labels modulo $d$. For $n=1$ the construction is trivial.
For a <primitive root of unity>, the <generalized Bell basis> obeys
$$
\langle\psi_{rs}|\psi_{r's'}\rangle
=\delta_{ss'}\frac1n\sum_{j=0}^{n-1}\omega^{j(r'-r)}
=\delta_{ss'}\delta_{rr'}.
$$
The last equality follows from the finite <geometric series>: a nonzero exponent difference modulo $n$ has sum zero, while zero difference has sum $n$. There are $n^2$ orthonormal vectors in an $n^2$-dimensional <Hilbert space>, so they form a complete basis.
For <qudit teleportation>, Alice's input is $|\chi\rangle_U=\sum_jc_j|j\rangle$. Alice and Bob share $|\psi_{00}\rangle_{aB}=n^{-1/2}\sum_k|k k\rangle$. Define the <qudit shift and phase operators> by $X|j\rangle=|j+1\rangle$ and $Z|j\rangle=\omega^j|j\rangle$. Alice measures $Ua$ in the <generalized Bell basis>. On outcome $(r,s)$, Bob's unnormalized state is
$$
\frac1n\sum_jc_j\omega^{-rj}|j+s\rangle_B
=\frac1nX^sZ^{-r}|\chi\rangle_B.
$$
All outcomes have probability $1/n^2$. Alice communicates $(r,s)$ and Bob applies
$$
\boxed{Z^rX^{-s},}
$$
which restores $|\chi\rangle$ exactly. The protocol consumes one maximally entangled pair of $n$-level systems, a local $n^2$-outcome measurement, and a classical message with $n^2$ possibilities. For a fixed-length binary encoding, $\lceil\log_2 n^2\rceil$ bits suffice. No measurement depends on the unknown amplitudes.
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