= Solution
Write the arbitrary joint <pure state> of the input $A$ and an external reference $R$ as
$$
|\Omega\rangle_{AR}=|0\rangle_A|r_0\rangle_R+|1\rangle_A|r_1\rangle_R,
\qquad \langle r_0|r_0\rangle+\langle r_1|r_1\rangle=1.
$$
The reference vectors need not be normalized or orthogonal. Share $|\Phi^+\rangle_{aB}$. Label Alice's <Bell states> by
$$
|\beta_{rs}\rangle_{Aa}
=\frac1{\sqrt2}\sum_{j=0}^1(-1)^{rj}|j\rangle_A|j\oplus s\rangle_a,
\qquad r,s\in\{0,1\}.
$$
Projecting onto this <Bell basis> gives the unnormalized state
$$
{}_{Aa}\langle\beta_{rs}|\left(|\Omega\rangle_{AR}|\Phi^+\rangle_{aB}\right)
=\frac12(X_B^sZ_B^r\otimes I_R)|\Omega\rangle_{BR}.
$$
Each outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies \b[$Z_B^rX_B^s$], reversing the two <Pauli gates>. The final state is exactly $|\Omega\rangle_{BR}$ for every outcome. In particular all input <entanglement> and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is <teleportation as an identity channel on a reference>.
Conditioned on the record, Alice's two measured carriers are in $|\beta_{rs}\rangle$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported <qubit> was initially <entangled> with a larger system.
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