= Solution
Let $\mathbf v_1=(c/2,0,0)$ and $\mathbf v_2=(c/(2\sqrt2),c/(2\sqrt2),0)$ in A. Each has <Lorentz factor> $g=2/\sqrt3$. The relative <Lorentz factor> of $S_2$ in the rest frame of $S_1$ follows from their <four-velocity> <tensor contraction>:
$$
\Gamma=-\frac{U_1\cdot U_2}{c^2}
=g^2\left(1-\frac{\mathbf v_1\cdot\mathbf v_2}{c^2}\right)
=\frac{8-\sqrt2}{6}.
$$
Let $b=\sqrt{1-\Gamma^{-2}}$ be the relative <speed> divided by $c$. In the rest frame of $S_1$, take $\theta$ between the <photon>'s propagation direction and $S_2$'s <velocity>. The observed <angular frequency> is the invariant $\omega=-k\cdot U$. Applying it to the emitter, whose rest <angular frequency> is $\omega_0$, gives $\omega_0=\Gamma\omega_1(1-b\cos\theta)$. Since $\omega_1=\omega_0$,
$$
\boxed{\cos\theta=\frac{1-\Gamma^{-1}}{b}
=\sqrt{\frac{\Gamma-1}{\Gamma+1}}
=\sqrt{\frac{2-\sqrt2}{14-\sqrt2}},\qquad \theta\simeq77.54^\circ.}
$$
This is a <relativistic Doppler effect> calculation using both emitters' and receivers' rest clocks; the time delay of reception introduces no extra factor for constant <velocities>.
For the direction in A, write the <wave four-vector> as $k^\mu=(\omega_A/c)(1,\mathbf n)$, where $\mathbf n$ is a <unit vector>. Equality of the emitted and received <angular frequencies> gives $k\cdot(U_1-U_2)=0$. Since the two laboratory <Lorentz factors> are equal, their time components cancel:
$$
\mathbf n\cdot(\mathbf v_1-\mathbf v_2)=0,
\qquad n_y=(\sqrt2-1)n_x.
$$
Thus the complete answer in three spatial dimensions is
$$
\boxed{\mathbf n=\cos\psi\,\big(\cos(\pi/8),\sin(\pi/8),0\big)+\sin\psi\,(0,0,1),\qquad 0\le\psi<2\pi.}
$$
If propagation is additionally assumed to lie in the stars' motion plane, this reduces to either direction along the internal <angle bisector>, $22.5^\circ$ or $202.5^\circ$ from $+x$. Reflection exchanging the stars' laboratory <velocities> explains the <angle bisector> by symmetry. \b[The source does not specify coplanar propagation or the stars' positions], so it does not fix the sign or exclude an out-of-plane ray. For example $\mathbf n=(0,0,1)$ gives $\omega_1=\omega_2=g\omega_A$ and satisfies the same no-shift condition. Initial positions can be chosen to make such a ray connect emission and reception. All these possibilities give the same rest-frame $\theta$ above.
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