= Solution
With signature $(+---)$ one convenient convention is
$$
F^{\mu\nu}=\begin{pmatrix}
0&-E_x/c&-E_y/c&-E_z/c\\E_x/c&0&-B_z&B_y\\E_y/c&B_z&0&-B_x\\E_z/c&-B_y&B_x&0
\end{pmatrix},
$$
which gives the stated dual after index lowering. The invariants are
$$
F^{\mu\nu}F_{\mu\nu}=2(B^2-E^2/c^2),
$$
$$
F^{\mu\nu}\widetilde F_{\mu\nu}=-4E\cdot B/c,\qquad
\widetilde F^{\mu\nu}\widetilde F_{\mu\nu}=2(E^2/c^2-B^2).
$$
For the boost,
$$
E'_y=\gamma(E_y-vB_z),\quad E'_z=\gamma(E_z+vB_y),
$$
$$
B'_y=\gamma(B_y+vE_z/c^2),\quad B'_z=\gamma(B_z-vE_y/c^2),
$$
with zero $x$ components.
In the final configuration, parallelism gives
$$
\sin\theta\,\beta^2-2\beta+\sin\theta=0,\qquad\beta=v/c.
$$
The physical root is
$$
\beta=\frac{1-\cos\theta}{\sin\theta}=\tan(\theta/2).
$$
\b[Thus $v\sim c\theta/2$ for small angle. As $\theta\to\pi/2$, $v\to c$, so exactly crossed equal-magnitude null fields cannot be aligned by any finite inertial boost.]
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