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For odd $n=2k+1$, a symmetric quantum teleportation resource can encode an unknown qubit into $E(\alpha |0⟩+\beta |1⟩)=\alpha |D_k^n⟩+\beta |D_{k+1}^n⟩$. Begin with $|D_{k+1}^{n+1}⟩$, apply a controlled-NOT gate from the input to the sender's resource qubit, apply a Pauli X gate to that resource qubit and a Hadamard gate to the input, and measure those two qubits. The four unnormalized remaining vectors are $E|\psi ⟩/2$, $EX|\psi ⟩/2$, $EZ|\psi ⟩/2$ and $EXZ|\psi ⟩/2$. Each outcome has probability $1/4$. Since $X^{\otimes n}$ interchanges $|D_k^n⟩$ and $|D_{k+1}^n⟩$, and $Z^{\otimes n}$ acts as logical $Z$ up to the common phase $(-1{)}^k$, local Pauli corrections make the output independent of the result. The one-qubit reduction of Dicke-state superpositions givesFor $n>1$ this distributes imperfect copies, as required by the no-cloning theorem. For $n=1$ it reduces to exact quantum teleportation.

The state in part (b) is symmetric in the receiving qubits. Single out Bob using the Dicke state recursion. For $k\ge 1$, the two vectors areDistinct excitation counts give orthogonal states of the other recipients. Taking their partial trace, the only surviving off-diagonal term comes from the common rest vector $|D_k^{N-1}⟩$. The one-qubit reduction of Dicke-state superpositions is consequentlyPut $p=|\alpha {|}^2$ and $q=|\beta {|}^2$, so $p+q=1$. The overlap with the input isFor $N=1$ there is no system to trace out, and the same matrix is $|\psi ⟩⟨\psi |$, with overlap one. For every $N>1$ the overlap is below one, even at its maximum $p=q=1/2$, so these are imperfect copies.