I working with an open, non-convex, simply-connected cone in $\mathbb{R}^{n \times n}$ with vertex at the origin, which is the set $\mathcal{S}_n$ of all real $n \times n$ matrices all of whose eigenvalues have negative real part (source). That the set is a cone means that if $x\in \mathcal{S}_n,$ then $tx \in \mathcal{S}_n$ for all $t\in \mathbb{R}_{>0}.$
I have observed the following symmetry:
Let $x_i$ denote the $i$th diagonal entry of a matrix in $\mathbb{R}^{n \times n},$ where $i\in\{1,2,\dots,n\}.$ Similarly, let $y_i$ denote the $i$th off-diagonal entry, where $i\in\{1,2,\dots,n(n-1)\}$ (how the entries are ordered turns out to not matter).
Now let $H(x_i,x_j)$ denote the plane spanned by $x_i$ and $x_j$ and let $H(y_i,y_j)$ be the same, but for diagonal entries.
Finally let $_xH_n^{ij}:=\mathcal{S}_n \bigcap H(x_i,x_j)$ and $_yH_n^{ij}:=\mathcal{S}_n \bigcap H(y_i,y_j).$ Then
$$_xH_n^{ij}={}_xH_n^{kl} \;\forall \;i,j,k,l$$ $$_yH_n^{ij}={}_yH_n^{kl} \;\forall \;i,j,k,l.$$
Informally: Say I have found a large number of matrices in $\mathcal{S}_n.$ If I then make a scatter-plot of all the values found for two (off-)diagonal entries, the result is the same no matter what pair of entries I chose.
How can this be explained and/or proven? Can it be stated in more concise/familiar terms?
At first I thought this was due to one being able to swap any two (off-)diagonal entries without it changing the eigenvalues, but this is only true for $n\leq2.$ $\mathcal{S}_n$ is invariant under similarity transforms - could this be the source of the symmetry? Per definition, the cone is also scale-invariant, but I doubt that this is relevant here.