= Solution
Take the labels $r,s\in\{0,\ldots,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega=1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00\rangle+|11\rangle)/\sqrt2$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. \b[The orthonormality claim is false for a nonprimitive root.]
For the intended result, choose a <primitive root of unity>, for example $\omega=e^{2\pi i/n}$. Since $|\omega|=1$, expansion in the <orthonormal basis> gives
$$
\begin{aligned}
\langle\psi_{rs}|\psi_{r's'}\rangle&=\frac1n\sum_{j,k=0}^{n-1}\omega^{-jr+kr'}\langle j|k\rangle\langle j+s|k+s'\rangle\\
&=\frac{\delta_{ss'}}n\sum_{j=0}^{n-1}\omega^{j(r'-r)}.
\end{aligned}
$$
If $r=r'$, the sum is $n$. If $r\ne r'$ modulo $n$, primitiveness gives $q=\omega^{r'-r}\ne1$ while $q^n=1$, and the <finite geometric series> is $(1-q^n)/(1-q)=0$. Thus
$$
\boxed{\langle\psi_{rs}|\psi_{r's'}\rangle=\delta_{rr'}\delta_{ss'}.}
$$
There are $n^2$ vectors in an $n^2$-dimensional <Hilbert space>, so they form the <generalized Bell basis>. Each has equal <Schmidt coefficients> $1/\sqrt n$ and is a <maximally entangled state>.
For <qudit teleportation>, let $C$ be Alice's unknown input, write $|\chi\rangle_C=\sum_j\alpha_j|j\rangle_C$, and let Alice and Bob share $|\Phi\rangle_{AB}=n^{-1/2}\sum_k|k\rangle_A|k\rangle_B$. Alice holds $C,A$ and Bob holds $B$. Define the <qudit shift and phase operators>
$$
X|j\rangle=|j+1\bmod n\rangle,\qquad Z|j\rangle=\omega^j|j\rangle.
$$
Both are unitary. Alice makes a joint <projective measurement> of $C,A$ in the <generalized Bell basis>, with projectors $|\psi_{rs}\rangle\langle\psi_{rs}|$. Completeness follows from the orthonormality just proved.
For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:
$$
\begin{aligned}
(\langle\psi_{rs}|_{CA}\otimes I_B)(|\chi\rangle_C\otimes|\Phi\rangle_{AB})&=\frac1n\sum_{j=0}^{n-1}\alpha_j\omega^{-jr}|j+s\rangle_B\\
&=\frac1n X^sZ^{-r}|\chi\rangle_B.
\end{aligned}
$$
The <Born rule> therefore gives $p_{rs}=1/n^2$, since $X^sZ^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^sZ^{-r}|\chi\rangle$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:
$$
\boxed{U_{rs}=Z^rX^{-s},\qquad U_{rs}X^sZ^{-r}|\chi\rangle=|\chi\rangle.}
$$
\b[The corrected receiver state is exactly the unknown input state for every outcome.] Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil\log_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2\log_2 n$. The protocol consumes the shared <maximally entangled state> and the original carrier is measured, so it does not retain an additional copy of the unknown state.
Equivalently, the complete joint <quantum state> has the teleportation expansion
$$
|\chi\rangle_C|\Phi\rangle_{AB}=\frac1n\sum_{r,s=0}^{n-1}|\psi_{rs}\rangle_{CA}\otimes X^sZ^{-r}|\chi\rangle_B.
$$
It follows from the contraction calculation and completeness of the <generalized Bell basis>. After correction the branch operators are all $I/n$, so their sum defines the identity <quantum channel>: $\sum_{r,s}(I/n)\rho(I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves <teleportation as an identity channel on a reference> and extends the protocol to mixed inputs.
Before receiving the label, Bob instead has the outcome-averaged state
$$
\rho_B=\frac1{n^2}\sum_{r,s}X^sZ^{-r}|\chi\rangle\langle\chi|Z^rX^{-s}=\frac In.
$$
To check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the <Heisenberg-Weyl twirling channel>. Thus the classical message is necessary to recover the input, consistently with the <no-communication theorem> derived in Question 1.
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