Frame $\mathcal{S}^{\prime}$ is moving with uniform speed $v$ in the $z$-direction relative to a laboratory frame $\mathcal{S}$. Using Cartesian coordinates and units such that $c=1$, the relevant Lorentz transformation is

$$t^{\prime}=\gamma(t-v z), \quad x^{\prime}=x, \quad y^{\prime}=y, \quad z^{\prime}=\gamma(z-v t)$$

where $\gamma=1 / \sqrt{1-v^{2}}$. A straight thin wire of infinite extent lies along the $z$-axis and carries charge and current line densities $\sigma$ and $J$ per unit length, as measured in $\mathcal{S}$. Stating carefully your assumptions show that the corresponding quantities in $\mathcal{S}^{\prime}$ are given by

$$\sigma^{\prime}=\gamma(\sigma-v J), \quad J^{\prime}=\gamma(J-v \sigma)$$

Using cylindrical polar coordinates, and the integral forms of the Maxwell equations $\nabla \cdot \mathbf{E}=\mu_{0} \rho$ and $\nabla \times \mathbf{B}=\mu_{0} \mathbf{j}$, derive the electric and magnetic fields outside the wire in both frames.

In a standard notation the Lorentz transformation for the electric and magnetic fields is

$$\mathbf{E}_{\|}^{\prime}=\mathbf{E}_{\|}, \quad \mathbf{B}_{\|}^{\prime}=\mathbf{B}_{\|}, \quad \quad \mathbf{E}_{\perp}^{\prime}=\gamma\left(\mathbf{E}_{\perp}+\mathbf{v} \times \mathbf{B}_{\perp}\right), \quad \mathbf{B}_{\perp}^{\prime}=\gamma\left(\mathbf{B}_{\perp}-\mathbf{v} \times \mathbf{E}_{\perp}\right)$$

Is your result consistent with this?