Let $H$ be a closed subgroup of a compact Lie group $G$. Suppose that $gHg^{-1}\subseteq H$ for some $g\in G$. Does it follow that $gHg^{-1}=H$?
If $H$ is connected the answer is clearly yes since $gHg^{-1}\subseteq H$ implies $\mathrm{Ad}(g)\mathfrak{h}\subseteq \mathfrak{h}$ which implies $\mathrm{Ad}(g)\mathfrak{h}=\mathfrak{h}$. But I'm not sure about the case $H$ is not connected.
(There are counterexamples for some infinite discrete groups, but those are not compact.)
Yes, it is true. Use the following proposition with $K=gHg^{-1}$.
Proposition. Let $H$ be a compact Lie group and $K$ a compact subgroup of $H$ of the same dimension and with the same number of connected components. Then, $K=H$.
Proof. Let $H_0$ be the connected component of the identity element of $H$. Then, $H/H_0$ is in bijection with the set of connected components of $H$. In particular, $H/H_0$ is a finite set (since $H$ is compact). The same statements hold for $K/K_0$. Since $\dim K=\dim H$, $K$ is open in $H$ and hence $K_0=H_0$. Thus, we have an injective map $K/K_0\to H/H_0$. Since $K/K_0$ and $H/H_0$ have the same number of elements, the map $K/K_0\to H/H_0$ is also surjective, and hence $H\subseteq K$. $\square$
