Let $n \in \mathbb{N}^{\ast}$ and $\mathrm{Sym}(n)$ (respectively $\mathrm{Spd}(n)$) denote the linear space (respectively set) of real $n \times n$ symmetric (respectively positive definite) matrices. I am interested in the following matrix differential equation :
$$ \frac{d\mathbf{X}}{dt} = \mathbf{V}\mathbf{S}^{-1} \mathbf{X}(t), \; t \in \mathbb{R} \tag{$\ast$} $$
where $\mathbf{V} \in \mathrm{Sym}(n)$, $\mathbf{S} \in \mathrm{Spd}(n)$ are known. The solutions of $(\ast)$ are of the form :
$$ t \in \mathbb{R}, \, \mathbf{X}(t) = \exp(t\mathbf{V}\mathbf{S}^{-1})\mathbf{X}(0). $$
My question is the following : if $\mathbf{X}(0) \in \mathrm{Sym}(n)$, does $(\star)$ have solutions in $\mathrm{Sym}(n)$ ? By that, I mean : is there a solution $\mathbf{X} : \mathbb{R} \to \mathrm{Sym}(n)$ of $(\ast$) ?
In your generality, no.
Let $V = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$ and $S^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$. Then
$$ VS^{-1} = \begin{pmatrix} 2& 1 \\ 2 & 1 \end{pmatrix} $$
Let $X(0) = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} $, the corresponding solution to the ODE is
$$ X(t) = \begin{pmatrix} e^{3t} & 1 \\ e^{3t} & -2 \end{pmatrix} $$
which is not symmetric.