Solution

We start analysing Rest frame S'

In the system S' the solenoid is at rest:

The solenoid is "long and flat": its turns define a plane; the magnetic moment $\mathbf{M}$
is perpendicular to that plane (normal direction). Let us take
take $\mathbf{M}' =M\,\hat{\mathbf{z}}$

There is no electric moment
$\mathbf{p}' = 0$.

In the laboratory system S, the solenoid moves with velocity $\mathbf{v} = v\,\hat{\mathbf{x}}$
parallel to the flat surfaces (hence, perpendicular to $\mathbf{M})$

Under a boost with velocity
$\mathbf{v}$
the electric dipole moment $\mathbf{p}$ and magnetic dipole moment $\mathbf{m}$ transform according to:

\begin{aligned} \mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\ \mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right). \end{aligned}

In our case:$\mathbf{p}' = 0$ and $\mathbf{v} \cdot \mathbf{m}' = 0$(because they are perpendicular)
Therefore:

$\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad \mathbf{m} = \gamma\,\mathbf{M}$

The magnetic moment in S increases by a factor of $\gamma$
and an electric moment appears perpendicular to $\mathbf{v}$ and to $\mathbf{M}$

For $v \ll c (\gamma \approx 1)$

In CGS: $\displaystyle \mathbf{p} \approx \frac{1}{c}\,\mathbf{v} \times \mathbf{M}$.

In SI: the factor is
$1/c^2$
$\displaystyle \mathbf{p} = \frac{1}{c^2}\,\mathbf{v} \times \mathbf{M}$

(with $\mathbf{M}$ in $A·m²$)

Magnitude (if $\mathbf{v} \perp \mathbf{M})$:
$p = \frac{v}{c}\,M (CGS)$
or $p = \frac{v}{c^2}\,M$(SI)

Direction:
perpendicular to$\mathbf{v}$and to $\mathbf{M}$,
that is, parallel to the flat surfaces but perpendicular to the motion.