$\newcommand\R{\Bbb R}$The answer to your both questions is yes, regardless of what the covariance matrix $\Sigma$ is.
Indeed, let $P_X:=X(X^\top X)^{-1}X^\top$, the orthoprojector onto $V_X:=X\R^d$, the column space of the random matrix $X$, so that $P_X\R^n=V_X$. Then for any orthogonal matrix $Q\in\R^{n\times n}$ we have $P_{QX}=QX(X^\top X)^{-1}X^\top Q^\top$ and hence $$V_{QX}=QX(X^\top X)^{-1}X^\top Q^\top\R^n =QX(X^\top X)^{-1}X^\top\R^n=QV_X.$$ Therefore and because $QX$ equals $X$ in distribution, it follows that $V_X$ equals $QV_X$ in distribution, for any orthogonal matrix $Q\in\R^{n\times n}$. So, $V_X$ is uniformly distributed. $\quad\Box$