Assume I have the following vector $X^*=(X_1,...,X_p,\tilde X_1, ...,\tilde X_p)$, with a data matrix $\mathbf{X^*}$ of dimension $n \times 2p$. The corresponding covariance matrix (lets define it with $G$), is given by
$G = \begin{bmatrix} \Sigma & \Sigma - \operatorname{diag}\{s\} \\ \Sigma - \operatorname{diag}\{s\} & \Sigma \end{bmatrix},$
with $\Sigma$ being a $p\times p$ symmetric covariance matrix of $(X_1,...,X_p)$ and also the symmetric covariance matrix of $(\tilde X_1, ...,\tilde X_p)$. The term $\operatorname{diag}\{s\}$ is simply a diagonal matrix with non-negative entries on the diagonals. Hence, $\Sigma - \operatorname{diag}\{s\}$ defines the covariances between $(X_1,...,X_p)$ and $(\tilde X_1, ...,\tilde X_p)$.
I want to show that for any column-wise permutation of $(X_1,...,X_p,\tilde X_1, ...,\tilde X_p)$ or more specifically the data matrix $\mathbf{X}^*$ encoded by the permutation matrix $P$, the covariance matrix $G$ does not change. With a column-wise permutation, I mean for example swapping the first and $p+1$th entry, which would be $(\tilde X_1, X_2 ..., X_p, X_1, \tilde X_2 ,...,\tilde X_p )$/ swapping the first and p+1th column in $\mathbf{X}^*$. This should have the same covariance matrix $G$. I know how to encode a column-wise permutation with a multiplication of permutation matrix $P$ from the right $\mathbf{X} P$ but I don't know how to prove such a statement in general.