Recover a set of vectors from a Gram matrix that have conjugate symmetry I have a set of complex column vectors concatenated in a matrix, $X$, for which I have the corresponding Gram matrix $G = X^*X$.  Because of the basis that I'm using (complex spherical harmonics), the vectors have a conjugate symmetry that can be expressed as $\overline{X} = MX$ where
$M = \begin{pmatrix}0 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 0
\end{pmatrix}\text{.}$
Otherwise put, if we label the rows of $X$ as $-1, 0, 1$ then $X_{m, n} = (-1)^m \overline{X_{-m, n}}$.  Note that this symmetry guarantees that the Gram matrix is real.
Now, suppose I'm given only the Gram matrix, $G$, and I want to recover any corresponding set of vectors $X$ that obey this conjugate symmetry, how can I do this?
I've tried a Cholesky decomposition $G = LL^{*}$, but this gives triangular matrices which clearly don't fulfill the conjugate symmetry.  Then I thought that maybe I can combine my symmetry criterion with a rotation of $L$ somehow, i.e. $G = LQQ^*L^*$, and find a $Q$ that would rotate $L$ into the complex plane such that the symmetry criterion is satisfied.  But I couldn't find a way to do this.
After attempting various decompositions of the Gram matrix I couldn't find a way that would give me a suitable set of vectors.  Can anybody help?

The ideas that you tried are pretty close.  For succinctness I assume $X$ has linearly independent columns.
You have $G=X^*X = \overline{X}^TX = (MX)^TX = X^TMX$.  So $G$ is Hermitian and symmetric and hence real.  Now run Cholesky over reals:
$G=LL^T$
then $X=QR$, with $R:=L^T$ and $Q$ is some unitary matrix that you may select.
$M(QR) = MX = \overline{X}= \overline{(QR)}= \overline{Q}R\implies MQ = \overline{Q}$
At this point, examining $M$ and $ MQ = \overline{Q}$ implies that row $2$ of $Q$ is real, and you just need to find rows 1 and 3 that are orthonormal to each other (and row 2) and are negative conjugates of each other.  So for example you may select,
$Q:= \begin{pmatrix}-\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\cdot i \\ 0 & 1 & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\cdot i
\end{pmatrix}\text{.}$
