Let's assume that i have a (real) $2N\times2N$ anti-symmetric matrix $B=\left\{ b_{ij}\right\} $ with the property that
$BB^{T}=\boldsymbol{1}$
where $\boldsymbol{1}$ is the identity matrix.
Is it true that I can always find an orthonormal matrix $U$ such that
$\tilde{B}=U^{T}BU$
is an anti-symmetric matrix with $\pm a$ on the counter diagonal, and zeros everywhere else? And if so, how to prove it?
For instance in the $4\times4$ case, if I start with the (almost) most general case
$B=\left(\begin{array}{cccc} 0 & b_{1} & b_{2} & b_{3}\\ -b_{1} & 0 & b_{3} & -b_{2}\\ -b_{2} & -b_{3} & 0 & b_{1}\\ -b_{3} & b_{2} & -b_{1} & 0 \end{array}\right)$
where $b_{1}^{2}+b_{2}^{2}+b_{3}^{2}=1$ by an appropriate choice of $U$ i can bring it to the form
$B=\left(\begin{array}{cccc} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0 \end{array}\right)$
What is true in general is that you can find an orthonormal matrix $P$ such that $\bar{B}=P^TBP$ is a block diagonal matrix with blocks of the form $M_b=\pmatrix{0 & b \cr -b & 0}$ (general theory of orthonormal reduction of normal matrices + some playing around the normal form using skewsymmetry).
Now $\bar{B}\bar{B}^T=P^TBP P^TB^TP=I_n,$ since $B$ and $P$ are orthonormal. Looking at the blocks, you see that you need $M_b M_b^T=I_2$, and that it implies $b^2=1$.
Hence the block involved are $M_1$ and $M_{-1}$. Changin gthe order of the vectors in your orthonormal basis (aka the columns of $P$) , you get the result you seek.