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The premise

It is known (doi:10.4153/CJM-1961-059-8) that given any skew-symmetric complex matrix $Z \in M_n(\mathbb{C})$ of rank $2k$, there is a unitary matrix $U$ such that $$ Z=UMU^\text{t}, $$ with $$ M= \begin{pmatrix} 0 & \lambda_1\\ -\lambda_1 & 0 \end{pmatrix} \oplus \begin{pmatrix} 0 & \lambda_2\\ -\lambda_2 & 0 \end{pmatrix} \oplus \dotsb\oplus \begin{pmatrix} 0 & \lambda_k\\ -\lambda_k & 0 \end{pmatrix} \oplus 0_{n-2k}, $$ where the $\lambda_a$ are the positive square roots of the eigenvalues of $Z^*Z$ and $0_{N}$ is the $N\times N$ zero matrix.

The question

What is our freedom in the choice of $U$? In other words, I would like to find the subgroup of the unitary group $\Sigma\subseteq \mathrm{U}(n)$ for which we have $$ UVMV^\text{t}U^\text{t} = UMU^\text{t},\quad \forall V\in\Sigma, $$ that is $$ \Sigma=\left\{ V\in\mathrm{U}(n) \,\mid\, V M V^\text{t} = M \right\}. $$ I know that when $\operatorname{rank}(Z)=2$ $$ \Sigma=\left\{ \begin{pmatrix} u & 0 \\ 0 &v \end{pmatrix} \,\middle|\, u \in \mathrm{SU}(2),v\in \mathrm{U}(n-2) \right\} \cong \mathrm{SU}(2) \times \mathrm{U}(n-2), $$ and that in general $ \mathrm{SU}(2)\times \mathrm{SU}(2)\times \dotsb \times \mathrm{SU}(2) \times\mathrm{U}(n-2k)$ (constructed in an analogous way) is a subgroup of $\Sigma$, but I am having trouble finding $\Sigma$ itself.

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