Let $(M,g)$ be a closed (compact without boundary) Riemannian manifold of finite dimension, with the volume measure $\mu:= \mu(E):=\int_{E}dvol_g \forall E \in \mathcal{B}(M),$ the Borel sigma algebra of $M.$

Let $G$ be a Lie group acting on $M$ by isometry, properly, but not necessarily freely. Under this action, denote by $O_p:=G.p,$ the $G$-orbit of $p.$ Fix $p^{*}\in M,$ and denote by $E$ the set $F:=\{p:\text{ there exists a unique minimizer of } d(p^{*}, ) \text{ in } O_p \}=\{p \in M: d(p^{*},p)=d(p^{*}, O_p), \text{ and } \forall r \ne p, r \in O_p, d(p^{*},r) > d(p^{*}, p)\}.$

My questions are:

(1) Is $\mu(M \setminus F):=dvol_g(M \setminus F)=0?$

(2) What is $M$ is not closed, but just complete? (still without boundary). Is the above measure of $F$ still zero?

P.S. This question is related to this question I asked earlier, and in some sense, its dual question, because that earlier question concerned itself with fixing the embedded submanifold $S$ and asking if the set of points from where the distance attains a unique minimizer to $S$ had measure zero. This one interchanges the role of the point and submanifold in question - here we fix the point $p^{*}$ and ask if the union of all $G$-orbits (that are embedded submanifolds of $M$) where the distance function from $p^{*}$ has a unique minimizer?