Proving or disproving that $M$ is a submanifold of the flag manifold of $G$ Let  $G$ be a compact connected Lie group and let $T$ be a maximal torus of $G$.  Denote by $\mathcal{F}=G/T$ the flag manifold of $G$. Let $\theta : G \rightarrow G$ be an involution on $G$. This involution induces an involution on the Lie algebra of $G$, that we denote also by $\theta$.
Consider the natural action of $G$ on $\mathcal{F}$. Let $x=gT \in \mathcal{F}$, denote by $G_x$ the stabilizer of $x$ in $G$ and denote by  $\mathfrak{g}_x$ its Lie algebra. (Note that $G_x =gTg^{-1}$).
Consider the set $M:= \lbrace x \in \mathcal{F}, \theta(\mathfrak{g}_x)=\mathfrak{g}_x \rbrace 
$.
$\textbf{Question}:$  Prove or disprove that $M$ is a submanifold of $\mathcal{F}$.
So far I didn't find any example for which $M$ is not a submanifold, and my thoughts are: Since the group $G$ acts transitively on the set $\tilde{\mathfrak{g}}:=\lbrace \mathfrak{g}_x , x \in \mathcal{F}\rbrace $ (this is because $\tilde{\mathfrak{g}}$ is the set of all cartan subalgebras of $\mathfrak{g}$) and since the set $M$ is a fixed point set of the involution $\theta$, then using this On the proof of that fixed point set of an involution is a submanifold), I conclude that $M$ is a submanifold. is this correct ?
$\textbf{Edit:}$ $T$ is assumed to be stable by $\theta$.
 A: $M$ is always a submanifold.
As you suggested, we can look at the space $\tilde{\mathfrak{g}}$.  However, $\tilde{\mathfrak{g}}$ is not diffeomorphic to $\mathcal{F}$.  Rather, it is a quotient of $\mathcal{F}$.
More specifically, we have:
Proposition:  Let $N = N_G(T)$ be the normalizer of $T$ in $G$.  Then $\tilde{\mathfrak{g}}$ is diffeomorphic to $G/N$.
Proof:  As you noted, $G$ acts transitively on $\tilde{\mathfrak{g}}$, so we only need to determine the stabilizer at a point.  Consider a point $\mathfrak{g}_x$ with $x\in T$, $g\in G$ stabilizes this set iff $Ad_g(\mathfrak{t}) = \mathfrak{t}$ (where $\mathfrak{t}$ denotes the Lie sub-algebra of $T$).  Exponentiating out, we find that conjugation by $g$ stablizes $T$, so $g\in N$.  Conversely, if $g\in N$, then $g$ stabilizes $T$, so $Ad_g$ stabilizes $\mathfrak{t}$.  $\square$.
Simply because $G/N$ is easier to type, I'll stop writing $\tilde{\mathfrak{g}}$ and instead right $G/N$ for the duration of this post.
Note that since $T\subseteq N$, we have a natural projection $\pi:G/T\rightarrow G/N$.  This projection is actually a covering with $|N/T|$ sheets.  (More generally, the projection is a fiber bundle with fiber $N/T$, but in the case of compact Lie groups, it is well-known that the Weyl group $N/T$ is finite.)
Proposition:  The $\theta$ action on $G$ induces a $\theta$ action on $G/N$.  Under the diffeomorphism $G/N\rightarrow \tilde{\mathfrak{g}}$ this $\theta$ action is equivariantly diffeomorphic to the $\theta$ action on $\tilde{\mathfrak{g}}$ given by $\mathfrak{g}_x\mapsto \theta(\mathfrak{g}_x)$.
Proof:  Define $\theta(gN) = \theta(g)N$.  To see this is well-defined, let $n\in N$.  Then $\theta(n) \in N_G(\theta(T)) = N_G(T) = N$, so $$\theta(gnN) = \theta(gn)N = \theta(g)\theta(n)N = \theta(g)N.$$
For the second statement, consider the diffeomorphism $\psi:G/N\rightarrow \tilde{\mathfrak{g}}$ given by $\psi(gN) = Ad_g \mathfrak{t}$.  Then $$\psi(\theta(gN)) = \psi(\theta(g)N) = Ad_{\theta(g)} \mathfrak{t} = Ad_{\theta(g)} \theta(\mathfrak{t}) = \theta(Ad_g \mathfrak{t}) = \theta(\psi(gN)).$$
$\square$
The induced $\theta$ action on $G/N$ is obviously an involution, so, as you already know, this implies $\mathrm{Fix}(\theta)\subseteq G/N$ is an embedded submanifold of $G/N$.
But we also obviously have that $M = \pi^{-1}(\mathrm{Fix}(\theta))$.  Thus, to finish off the proof, we need only note that the inverse image of a submanifold under a covering is itself a submanifold.  Indeed, slice charts on $\mathrm{Fix}(\theta)$ pull back to to slice charts on $M$.
