A real skew-symmetric matrix $A$ can be diagonalized with complex eigenvectors and pure imaginary eigenvalues:
$$A=V S V^*$$
where $S$ is:
$$S = \begin{pmatrix} -\lambda_1\mathrm{i} & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_1\mathrm{i} & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & -\lambda_2\mathrm{i} & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2\mathrm{i} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}$$ all $\lambda_i$ are real and positive, and $V$ is a complex unitary matrix.
Similarly, $A$ can be real-Schur-decomposed, with both real Schur form and vectors, i.e.:
$$A = U \Sigma U^\mathrm{T}$$
with $\Sigma$ given by the same $\lambda_i$’s:
$$\Sigma = \begin{pmatrix} 0 & \lambda_1 & 0 & 0 & \cdots & 0 & 0 \\ -\lambda_1 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2 & \cdots & 0 & 0 \\ 0 & 0 & -\lambda_2 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}$$ and $U$ a real unitary (orthogonal) matrix.
Is there any relationship between $V$ and $U$. Specifically, given $V$ (and $S$), is there an “easy” way to get $U$ (or vice-versa, $U,\Sigma\to V$)?
Yes there is. Denote $$ D = \pmatrix{-i & 0\\ 0 & i}, \quad J = \pmatrix{0 & 1\\ -1 & 0}, \quad W = \frac 1{\sqrt{2}}\pmatrix{i & 1\\1 & i}. $$ We find that $W^*JW = D$. Consequently, if $$ \Sigma = \pmatrix{\lambda_1 J \\ & \ddots \\ && \lambda_k J\\ &&& 0}, $$ Then we have $\Omega^* \Sigma \Omega = S$, where $$ \Omega = \pmatrix{W\\ & \ddots \\ && W\\ &&& I}. $$ Now, if $A = VSV^*$, then $A = V\Omega^* \Sigma \Omega V^* = (V\Omega^*)\Sigma (V \Omega^*)^*$. In other words, if we are given $V$, then $U = V\Omega^*$. Conversely, if we are given $U$, then $V = U\Omega$.