= Solution
The conserved initial <four-momentum> of a particle with positive <rest mass> satisfies $P_\mu P^\mu=m^2c^2>0$. A single <photon> instead has null <four-momentum>, $K_\mu K^\mu=0$. <Four-momentum conservation> for an isolated one-body decay would require $P^\mu=K^\mu$, contradicting these <Lorentz invariants>. Thus \b[decay into a single photon is impossible in every inertial frame].
Equivalently, the incoming <energy> and <momentum> obey $E^2-c^2|\mathbf p|^2=m^2c^4$, whereas a single outgoing <photon> obeys $E=c|\mathbf p|$. Changing the laboratory <speed> cannot change this mismatch of the invariant <mass shell>.
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