Is this a submanifold? 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:
$$\psi : G \times TM \to TM$$
$$\psi(g,p,X) := (\rho(g)p,d\rho(g)X).$$
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 $g\in G$ consider $\eta_g(p,X) \equiv \eta(g,p,X) := d^2_{TM}\left((p,X),\psi(g,p,X)\right).$ Then, one has:
    $$\eta_g^{-1}(0) = \{(p,X) :(\rho(g)p,d\rho(g)X) = (p,X)\}.$$
    So,
    $$\tilde S = \bigcup_{g\in G}\eta_g^{-1}(0). $$
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 $T_pG\cdot p$ 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.
 A: Your definition implies that
$$ \tilde S = \bigcup_{p\in M} T_pM^{G_p}. $$
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 $p\in M$.
$G$ 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 $T(S^2\setminus\{n,s\})\cup \{n,s\}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
EDIT
To address the modified question: If $p\in S$, then the orbit $Gp$ has a neighborhood homeomorphic to $G\times_{G_p} \mathcal H_p$ and we're assuming that $\mathcal H_p^{G_p} \neq 0$. If $q\in \mathcal H_p$ 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 $M$.
A: For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ 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 $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ 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.
