= Solution
For a vacuum <electromagnetic wave>, the <four-wavevector> is $K^\mu=(\omega/c,\mathbf k)$ with $|\mathbf k|=\omega/c$ and $K^2=0$. Its phase can be written
$$
\varphi=K_\mu X^\mu+\varphi_0=\omega t-\mathbf k\cdot\mathbf x+\varphi_0.
$$
Because $K$ and $X$ transform as <Lorentz four-vectors>, their contraction is a <Lorentz scalar>. This proves phase invariance, with the same phase convention in both frames.
For a hypothetical free emission $e\to e+\gamma$, <four-momentum conservation> gives $P=P'+Q$. The initial and final <electrons> have $P^2=P'^2=m_e^2c^2$, and $Q^2=0$. Squaring gives $P'\cdot Q=0$. In the final <electron>'s rest frame, however, $P'\cdot Q=m_eE_\gamma>0$ for a nonzero future-directed <photon>. Thus
$$
\boxed{\text{an isolated on-shell electron cannot emit a real photon in vacuum}.}
$$
An external field, another body or a medium can exchange <momentum> and change this conclusion; those are different systems.
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