$\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}\DeclareMathOperator\SU{SU}\DeclareMathOperator\Lift{Lift}$The subgroup of $ \SO(n) $ of determinant-$1$ signed permutations has order $ n!2^n/2 $. I will call this group $ W_n $.
Question:
Is it true that $ W_n $ is a maximal proper closed subgroup of $ SO(n) $ if and only if $ n $ is not a power of $ 2 $?
Context: $$ W_3 \cong S_4 $$ is isomorphic to the symmetric group on 4 letters. It is a maximal proper closed subgroup of $ \SO(3) $.
It is also true that there are no closed subgroups of $ \SO(5) $ containing $ W_5 $ and none in $ \SO(6) $ containing $ W_6 $.
$$ W_2 \cong C_4 $$ is the cyclic group of order 4. It is not maximal. The infinitely many cyclic subgroups of order $ 4m $ all contain $ W_2 $.
And for $ W_4 $ we have a chain of strict containments $$ W_4 \subsetneq \Lift(S_4 \times S_4) \subsetneq \SO(4) $$ where $\Lift$ denotes the Lift through the double cover $ \SO(4) \to \SO(3) \times \SO(3) $.
I think that $ W_n $ fails to be maximal for every $ n=2^k $ a power of $ 2 $. Indeed it seems to me that $ W_{2^k} $ normalizes a certain extraspecial subgroup $ E \subset SO(2^k) $ of order $ 2^{2k+1} $. The full normalizer of $ E $ is generated by the signed permutations together with tensor products involving $$ \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$