= Solution
In the instantaneous rest frame the total radiation has zero net three-momentum, so its emitted four-momentum is proportional to the <four-velocity>. A Lorentz boost multiplies its energy by $\gamma$, while the elapsed coordinate time for the source is also multiplied by $\gamma$. Their ratio, total instantaneous radiated power, is therefore invariant. This concerns total emitted power, not the directional flux or an observer's photon arrival rate.
Use the metric of signature $(+---)$ and dots for proper-time derivatives. From $\gamma=(1-v^2)^{-1/2}$ obtain $\dot\gamma=\gamma^3\alpha$, where $\alpha=\mathbf v\cdot\dot{\mathbf v}$. Differentiating $U=(\gamma,\gamma\mathbf v)$ gives
$$
\boxed{\dot U=(\gamma^3\alpha,\gamma^3\alpha\mathbf v+\gamma\dot{\mathbf v}).}
$$
Squaring in the <Minkowski metric>, using $1-v^2=\gamma^{-2}$, gives $\dot U\cdot\dot U=\gamma^6\alpha^2(1-v^2)-2\gamma^4\alpha^2-\gamma^2|\dot{\mathbf v}|^2$, hence
$$
\boxed{\dot U\cdot\dot U=-\gamma^4\alpha^2-\gamma^2|\dot{\mathbf v}|^2.}
$$
In the rest frame, $-\dot U^2$ is the squared ordinary acceleration, so invariance of power and the low-speed <Larmor formula> give $P=-(\mu_0q^2/6\pi)\dot U^2$. In terms of coordinate-time acceleration $\mathbf a=d\mathbf v/dt$, this is the <Lienard formula>
$$
\boxed{P=\frac{\mu_0q^2}{6\pi}\gamma^6\bigl(|\mathbf a|^2-|\mathbf v\times\mathbf a|^2\bigr),\qquad c=1.}
$$
For circular motion perpendicular to $\mathbf B$, $\mathbf a\perp\mathbf v$ and the <Lorentz force> gives $\gamma m|\mathbf a|=|q|v|\mathbf B|$. Thus
$$
\boxed{P=\frac{\mu_0q^4}{6\pi m^2}(\gamma^2-1)|\mathbf B|^2.}
$$
At low speed $\gamma^2-1\sim v^2$, reproducing ordinary Larmor radiation with magnetic acceleration $|q|vB/m$.
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