= Solution
Assume an infinite, stationary, uniform thin wire, and no added external homogeneous fields. Its <four-current> is $j^\mu=(\sigma,0,0,J)\delta(x)\delta(y)$. A boost along the wire leaves the transverse delta functions unchanged and transforms the time and longitudinal components, giving
$$
\boxed{\sigma'=\gamma(\sigma-vJ),\qquad J'=\gamma(J-v\sigma).}
$$
The densities are measured per unit simultaneous length in their respective frames; treating charge alone as invariant per unchanged coordinate length would miss the current contribution.
For a coaxial cylinder, <Gauss's law> gives $E_r(2\pi rL)=\mu_0\sigma L$ in the stated $c=1$ units. For a circular loop, <Ampère's law> gives $B_\varphi2\pi r=\mu_0J$. Thus $E_r=\mu_0\sigma/(2\pi r)$ and $B_\varphi=\mu_0J/(2\pi r)$. The same reasoning in the primed frame gives the same expressions with primed densities.
Since $\widehat z\times\widehat\varphi=-\widehat r$ and $\widehat z\times\widehat r=\widehat\varphi$, the <Lorentz transformation of electromagnetic fields> gives $E'_r=\gamma(E_r-vB_\varphi)$ and $B'_\varphi=\gamma(B_\varphi-vE_r)$. Substituting the laboratory fields reproduces the primed-density formulas exactly; parallel components vanish in both frames.
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