Let $\mathfrak{g}$ be a semisimple complex Lie algebra, and $G$ the adjoint group of $\mathfrak{g}$. Let $\sigma:\mathfrak{g}\rightarrow \mathfrak{g}$ be a complex conjugation and $\mathfrak{g}^{\sigma}$ the corresponding real form of $\mathfrak{g}$. We also consider the real form $G^{\sigma}$ of $G$, here $\sigma:G\rightarrow G$ is a lift of $\sigma:\mathfrak{g}\rightarrow \mathfrak{g}$ to $G$. At some references $G^{\sigma}$ is called the group of "quasi-inner" automorphisms. Let $K^{\sigma}$ be the maximal compact subgroup of $G^{\sigma}$. It is known that the connected component groups of $K^{\sigma}$ and $G^{\sigma}$ are the same. Now let $\Pi_0$ be the set of simple compact roots of $\mathfrak{g}^{\sigma}$ (i.e. the black dots on the Satake diagram of $\mathfrak{g}^{\sigma}$). Let us consider a minimal $\sigma$-stable parabolic subgroup $P_{\Pi_0}\subset G$. Let $M_{\Pi_0}\subset G$ be the Levi subgroup of $P_{\Pi_0}$. Let $\mathfrak{m}_{\Pi_0}=Lie M_{\Pi_0}$. Let $\mathfrak{g}_{\Pi_0}=[\mathfrak{m}_{\Pi_0},\mathfrak{m}_{\Pi_0}]$ be the semisimple subalgebra of $\mathfrak{g}$ with the set of simple roots $\Pi_0$, and $G_{\Pi_0}$ the adjoint group of $\mathfrak{g}_{\Pi_0}$ considered as a subgroup of $G$. It is easy to see that $G_{\Pi_0}$ and $M_{\Pi_0}$ are $\sigma$-stable. Moreover, the set of $\sigma$-fixed points in $G_{\Pi_0}$ is a compact and connected subgroup of $G^{\sigma}$.

Is the above set up right? What is the relationship between the subgroups of $\sigma$-fixed points $K^{\sigma}$, $G_{\Pi_0}^{\sigma}$ and $M_{\Pi_0}^{\sigma}$?

In particular, let $\mathfrak{g}^{\sigma}$ be of real rank $1$. It is known that $M_{\Pi_0}^{\sigma}$ acts by inner automorphisms on $\mathfrak{g}^{\sigma}$. How the groups of automorphisms induced by $K^{\sigma}$, $G_{\Pi_0}^{\sigma}$ and $M_{\Pi_0}^{\sigma}$ on $\mathfrak{g}^{\sigma}$ are related?