Parabolic subalgebra conjugate under Weyl group Let $G$ be a semisimple (or reductive (using the definition of Knapp)) Lie group. Then any two minimal parabolic subalgebras resp. subgroups are conjugated under $\operatorname{Ad}(K)$ resp. $K$ ($K$ being maximally compact in $G$).
My question is: Is it sufficient to take $k\in K$ in the normalizer $N_K(\mathfrak a)$ of $\mathfrak a$ in $K$ (resp. the Weyl group $N_K(\mathfrak a)/Z_K(\mathfrak a)$), so that any two conjugated parabolics differ only be another choice of positive system in $\mathfrak a$? Here, $\mathfrak a$ denotes a maximal abelian subspace of $\mathfrak p$, the $-1$-eigenspace of the Cartan decomposition of $\mathfrak g=\mathfrak k\oplus\mathfrak p$.
 A: It is not sufficient.
Example:
$G= SL_2(\mathbb R)$, $\mathfrak g = \mathfrak{sl}_2(\mathbb R)$,
$\mathfrak k = \mathfrak{so}_2(\mathbb R) = \{ \pmatrix{0&\theta \\ -\theta & 0} \vert \theta \in \mathbb R \}$, $\mathfrak p = \{ \pmatrix{a & c \\ c & -a} \vert \; a,c \in \mathbb R \}$,
$K= SO_2(\mathbb R) = \{ R_\theta := \pmatrix{\cos(\theta) & \sin(\theta) \\ -\sin(\theta) &\cos(\theta) } \vert \;\theta \in \mathbb R \}$.
Now the first problem is that there are various choices for maximal abelian subspaces inside $\mathfrak p$ -- actually, each one-dimensional subspace of $\mathfrak p$ is one. The standard choice is
$$\mathfrak a := \{ \pmatrix{a & 0 \\ 0 & -a} \vert \; a \in \mathbb R \},$$
but
$$\mathfrak c := \{ \pmatrix{0 & c \\ c & 0} \vert \; c \in \mathbb R \}$$
theoretically is not better or worse. Now a well-known minimal parabolic (= Borel) subalgebra of $\mathfrak g$ is
$$\mathfrak b_a := \{ \pmatrix{a & b \\ 0 & -a} \vert \; a, b \in \mathbb R \}$$
which contains $\mathfrak a$; but another Borel subalgebra is
$$\mathfrak b_c := \{ \pmatrix{r & c+r \\ c-r & -r} \vert \; c,r \in \mathbb R \}$$
which contains $\mathfrak c$. And you can check that $\mathfrak b_a$ and $\mathfrak b_c$ are indeed conjugate via $K$ as predicted by the result quoted in your first paragraph, namely, with $R_{\pi/4} = \dfrac12 \sqrt{2} \cdot \pmatrix{1 & 1 \\ -1 & 1} \in K $ we have
$$ R_{\pi/4} \cdot \mathfrak b_a \cdot R_{\pi/4}^{-1} = \mathfrak b_c$$
including
$$ R_{\pi/4} \cdot \mathfrak a \cdot R_{\pi/4}^{-1} = \mathfrak c$$
i.e. $R_{\pi/4} \not \in N_K(\mathfrak a)$, and of course nothing in $N_K(\mathfrak a)$ could possibly conjugate $\mathfrak b_a$ and $\mathfrak b_c$ (as $\mathfrak a \cap \mathfrak b_c = \{0\}$).
Indeed $N_K(\mathfrak a)$ contains just four elements, $ \pm \pmatrix{1 & 0 \\ 0 & 1}$ and $\pm \pmatrix{0 & 1 \\ -1 & 0}$; the first two centralize $\mathfrak a$, so the Weyl group here indeed just has two elements, the non-trivial one being represented by $R_{\pi/2} = \pmatrix{0 & 1 \\ -1 & 0}$, and this conjugates $\mathfrak b_a$ and
$$\mathfrak b_a^- := \{ \pmatrix{a & 0 \\ b & -a} \vert \; a, b \in \mathbb R \}$$
which seem to be the only two minimal parabolics which contain $\mathfrak a$, and these two really correspond just to the other choice of the one positive root in the root system to $\mathfrak a$. But since you asked about general minimal parabolics ...
