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Let $(M,g)$ be a compact Riemannian manifold with an isometric action $rho : G to mathrm{Iso}(M)$ by a compact Lie group $G$. There is a natural extension of $rho$ to $TM$ given by:
$$psi : G times TM to TM$$
$$psi(g,p,X) := (rho(g)p,drho(g)X).$$

I would like to know if the set:
$$TM supset tilde S := {(p,X) in TM : G_X = G_p}$$ is a submanifold of $TM$, and further, if $pi : TM to M$ is the natural projection, then $pi(tilde S)$ is a submanifold of $M$.

I tried the following approach:

For each $gin G$ consider $eta_g(p,X) equiv eta(g,p,X) := d^2_{TM}left((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^{-1}(0) = {(p,X) :(rho(g)p,drho(g)X) = (p,X)}.$$
So,
$$tilde S = bigcup_{gin G}eta_g^{-1}(0). $$

But according to my calculation $0$ is not a regular value of $eta_g$.

I appreciate any help.

EDIT

Thanks to all the answers and comments. I shall change a little the candidate to manifold to another that will be more helpful to me.

I would like to know if one denotes by $cal H_p$ the orthogonal complement to $T_pGcdot p$ on the $g$-metric, then the set
$$S := {p in M : exists X in mathcal H_p : G_X = G_p}$$
is a submanifold of $M$. In fact, this is what I was trying to prove at first.

Your definition implies that
$$ tilde S = bigcup_{pin M} T_pM^{G_p}. $$
In particular, $pi(tilde S) = M$, and $tilde S$ will be a submanifold of $TM$ iff the dimension of $T_pM^{G_p}$ is the same for all $pin M$.

$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $tilde S$ is then $T(S^2setminus{n,s})cup {n,s}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.

EDIT

To address the modified question: If $pin S$, then the orbit $Gp$ has a neighborhood homeomorphic to $Gtimes_{G_p} mathcal H_p$ and we're assuming that $mathcal H_p^{G_p} neq 0$. If $qin mathcal H_p$ is any point, then $mathcal H_p^{G_p}$ will be fixed by $G_q$ and still be orthogonal to $Gq$, hence $q$ and every point in $Gq$ will be in $S$. This shows that a neighborhood of $Gp$ is contained in $S$. Thus $S$ is an open submanifold of $M$.

For the obvious reflection action of $mathbb{Z}/2mathbb{Z}$ on $S^1$ this set $tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$.
Then $tilde S$ is $T(S^1setminus {p,q}) cup {p_0,q_0}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.