= Solution
Combining the real transformations gives
$$
F'_x=F_x,\qquad F'_y=\gamma(F_y+ivF_z),\qquad F'_z=\gamma(F_z-ivF_y).
$$
Let $v=\tanh\psi$, so $\gamma=\cosh\psi$ and $\gamma v=\sinh\psi$. The usual rotation about the positive $x$-axis has the $yz$ block $\left(\begin{smallmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{smallmatrix}\right)$. Taking $\boxed{n=(1,0,0),\ \theta=-i\psi}$ gives exactly the displayed transformation, because $\cos(-i\psi)=\cosh\psi$ and $\sin(-i\psi)=-i\sinh\psi$. This is a complex rotation, not a real spatial rotation.
Complex rotations preserve the bilinear dot product, without conjugation. Hence
$$
F\cdot F=E^2-B^2+2iE\cdot B
$$
is invariant, giving the two independent real <Lorentz invariants>
$$
\boxed{E^2-B^2,\qquad E\cdot B.}
$$
Their completeness refers to scalar invariants of the full <Lorentz group>. For a generic non-null field, a boost along $E\times B$ makes the fields parallel. Indeed with $s=|E\times B|$ and $A=E^2+B^2$, the required speed solves $s(1+v^2)-Av=0$; the smaller root is $v=[A-\sqrt{(E^2-B^2)^2+4(E\cdot B)^2}]/(2s)<1$. For $s=0$ the fields are already parallel. A real rotation then gives $E=e n$, $B=b n$, leaving just $e^2-b^2$ and $eb$ as independent scalar parameters. Thus there are precisely two independent continuous scalar invariants on generic orbits; null fields have special orbits and the invariants do not distinguish zero from a nonzero null field.
If the printed phrase is read as invariance under only this single-axis boost, it admits extra fixed components $F_x$, so the assertion of precisely two must be understood in its Lorentz-scalar sense.
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