= Solution
The <J gate> is $J(\alpha)=HP(\alpha)$, with $P(\alpha)=\operatorname{diag}(1,e^{i\alpha})$. Prepare a fresh <qubit> in $|+\rangle$ and apply the <Controlled-Z gate> $E$ between it and the input $|\psi\rangle=u|0\rangle+v|1\rangle$. The resulting state is
$$
u|0\rangle|+\rangle+v|1\rangle|-\rangle.
$$
Measure the input in the <equatorial qubit measurement> basis $|\alpha_s\rangle=(|0\rangle+(-1)^se^{-i\alpha}|1\rangle)/\sqrt2$. The unnormalized output is
$$
\frac1{\sqrt2}\left(u|+\rangle+(-1)^se^{i\alpha}v|-\rangle\right)
=\frac1{\sqrt2}X^sJ(\alpha)|\psi\rangle.
$$
Each outcome has probability $1/2$. Thus <one-bit teleportation> realizes
$$
\boxed{|\psi\rangle\longmapsto X^sJ(\alpha)|\psi\rangle}
$$
on the new <qubit>. Apply the known <Pauli X gate> correction $X^s$ for the literal $J(\alpha)$ output, or keep the correction in a <Pauli frame> and adapt later measurements. The old <qubit> is measured, so this is not cloning the input.
Direct multiplication of the given matrices gives $J(\alpha)X^s=e^{+is\alpha}Z^sJ((-1)^s\alpha)$. The displayed negative exponent in the supplied relation has the wrong sign for exact matrix equality. The discrepancy is only a <global phase> in a fixed measurement branch, so it does not change this measurement implementation or its outcome probabilities. The positive-sign identity is used when tracking exact matrices.
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