= Solution
The <spinor> transformation is $\chi'=S\chi$, with $S=\operatorname{diag}(e^{\kappa/2},e^{-\kappa/2})$. Its squared norm becomes
$$
\mathcal R=e^\kappa\cos^2(\theta/2)+e^{-\kappa}\sin^2(\theta/2)
=\cosh\kappa+\sinh\kappa\cos\theta.
$$
Dividing by $\sqrt{\mathcal R}$ isolates the new direction without discarding the physical scale factor in $X'$. The lower-to-upper component ratio gives
$$
\boxed{\tan\frac{\theta'}2=e^{-\kappa}\tan\frac\theta2,\qquad
\phi'=\phi\pmod{2\pi}.}
$$
Equivalently,
$$
\boxed{\cos\theta'=\frac{\cos\theta+\tanh\kappa}
{1+\tanh\kappa\cos\theta},\qquad
\sin\theta'=\frac{\sin\theta}{\mathcal R}.}
$$
This is <relativistic aberration> for the active positive-$z$ boost specified in the question. At the poles, <azimuth> is undefined rather than a measurable transformed variable. Under the literal <matrix> convention from (c), the actual <azimuth> $-\phi$ is likewise unchanged. The <time> coordinate is multiplied by $\mathcal R$; normalizing a <spinor> for its direction must not erase that boost-induced change of null-vector scale.
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