= Solution
Use rationalized electromagnetic units with $c=1$ and electrostatic constant $1/(4\pi)$. The <four-potential> $A^a=(\varphi,A_x,A_y,A_z)$ transforms by the same <matrix> as the coordinates:
$$
\boxed{\varphi'=\gamma(\varphi-vA_x),\quad A_x'=\gamma(A_x-v\varphi),\quad A_y'=A_y,\quad A_z'=A_z.}
$$
Write $r=\sqrt{y^2+z^2}$ and $\mathbf e_r=(0,y,z)/r$. <Gauss's law> for the line at rest gives $\mathbf E'=\sigma\mathbf e_r/(2\pi r)$, $\mathbf B'=0$. A suitable potential is $\varphi'=-\sigma\log(r/r_*)/(2\pi)$, $\mathbf A'=0$. The inverse <Lorentz transformation> gives $\varphi=\gamma\varphi'$, $A_x=\gamma v\varphi'$. The fields from $\mathbf E=-\nabla\varphi-\partial_t\mathbf A$ and $\mathbf B=\nabla\times\mathbf A$ are therefore
$$
\boxed{\mathbf E=\frac{\gamma\sigma}{2\pi r}\mathbf e_r,\qquad\mathbf B=\frac{\gamma v\sigma}{2\pi r}\mathbf e_x\times\mathbf e_r.}
$$
The sign agrees with a <line charge> moving in the positive $x$ direction: its density is $\gamma\sigma$ and its current $\gamma v\sigma$. In SI units insert $1/\epsilon_0$ in the electric expressions and replace $\mathbf B=\mathbf v\times\mathbf E$ by $\mathbf B=\mathbf v\times\mathbf E/c^2$.
The two invariants check explicitly:
$$
\mathbf E\cdot\mathbf B=0=\mathbf E'\cdot\mathbf B',\qquad E^2-B^2=\gamma^2(1-v^2)\frac{\sigma^2}{4\pi^2r^2}=E'^2-B'^2.
$$
For $|v|\ll1$, the <electric field> differs from the static field only at order $v^2$, while the <magnetic field> is of order $v$ and is the field of the moving line current. These fields are evaluated off the charged axis.
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