= Solution
Take the wire along the $x$-axis and let $r$ be perpendicular distance from it. <Gauss's law> and cylindrical symmetry give
$$
\mathbf E=\frac{\lambda}{2\pi\varepsilon_0r}\,\mathbf e_r,
\qquad \mathbf B=0.
$$
For an observer boosted parallel to the wire,
$$
\boxed{
\mathbf E'=\gamma\frac{\lambda}{2\pi\varepsilon_0r}\,\mathbf e_r,
\qquad
\mathbf B'=-\frac{\gamma}{c^2}\mathbf v\times\mathbf E}.
$$
The moving observer sees length contraction and hence line density $\lambda'=\gamma\lambda$, together with current $I'=-\gamma\lambda v$. The transformed magnetic field is exactly
$$
B'=\frac{\mu_0|I'|}{2\pi r},
$$
with the direction prescribed by the current. This is <Ampère-Maxwell equation>[Ampère's law] for the current seen in the moving frame.
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