= Solution
Assume homogeneous space and time, isotropic space, inertial frames related linearly, the relativity principle and a common isotropic speed $c$ for light. Take coincident origins at zero and parallel axes, with $S'$ moving at $v$ along $x$. The primed origin requires $x'=A(x-vt)$. Write $t'=Bx+Dt$; mapping each light ray $x=\pm ct$ into $x'=\pm ct'$ gives $D=A$ and $B=-Av/c^2$. Rotational symmetry about the boost axis gives a common transverse scale $B(v)$; reversing the boost direction by spatial isotropy gives $B(v)=B(-v)$. Reciprocity then gives $B(v)^2=1$, and continuity and the common orientation select $B(v)=1$. Thus transverse directions are unchanged. Reciprocity requires the inverse to have the same scale $A$ with $v$ replaced by $-v$. Composing gives $A^2(1-v^2/c^2)=1$; continuity from $A=1$ at $v=0$ selects the positive root. Hence the <Lorentz transformation> is
$$
\boxed{x'=\gamma(x-vt),\quad t'=\gamma(t-vx/c^2),\quad y'=y,\quad z'=z,\qquad \gamma=(1-v^2/c^2)^{-1/2}.}
$$
The printed wavelength formula uses the angle of the line of sight from observer towards star, opposite to photon propagation. Choose the transverse axis so the ray lies in the $xy$ plane. With that directed-angle convention the photon <four-momentum> is
$$
\boxed{p'^\mu=\frac{h\nu'}c(1,-\cos\theta',-\sin\theta',0),\qquad p^\mu=\frac{h\nu}c(1,-\cos\theta,-\sin\theta,0).}
$$
The inverse coordinate boost takes a rest-frame four-vector to $S$, giving $p^0=\gamma(p'^0+\beta p'^1)$ and $p^1=\gamma(p'^1+\beta p'^0)$, where $\beta=v/c$. Thus $\nu=\gamma\nu'(1-\beta\cos\theta')$, and the <aberration of light> relation is $\cos\theta=(\cos\theta'-\beta)/(1-\beta\cos\theta')$. More directly, the boost from $S$ to $S'$ gives $p'^0=\gamma(p^0-\beta p^1)$, hence $\nu'=\gamma\nu(1+\beta\cos\theta)$. Since wavelength is $c/\nu$, the <relativistic Doppler effect> gives
$$
\boxed{\lambda=\lambda'\gamma(1+\beta\cos\theta).}
$$
If the angle were instead measured along the actual photon momentum, both cosine signs would reverse and the same physical relation would have a minus sign. Specifying this orientation avoids confusing the two versions.
For $\cos\theta=1$, $\lambda/\lambda'=\sqrt{(1+\beta)/(1-\beta)}$: a receding star is redshifted. For $\cos\theta=-1$, the ratio is $\sqrt{(1-\beta)/(1+\beta)}$: an approaching star is blueshifted. For $\cos\theta=0$, $\boxed{\lambda/\lambda'=\gamma}$: the transverse redshift is the time-dilation effect. This transverse condition refers to the observer's angle, not $\theta'=\pi/2$ in the source frame.
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