**Question**: Consider the complex simple Lie group $E_6$. Let $\lambda_1$ and $\lambda_6$ be the fundamental weights defining the $27$-dimensional representation $V$ and $V^*$, resp. Consider the complex span $\mathfrak a^*=\langle \lambda_1,\lambda_6\rangle$. The subquotient of the Weyl group $W=W(E_6)$ that acts
on $\mathfrak a^*$ is $\Gamma:=N_W(\mathfrak a^*)/Z_W(\mathfrak a^*)=A_2$.

If two elements in $\mathfrak a^*$ are $W$-conjugate, are they $\Gamma$-conjugate? Equivalently, does $$W\cdot\lambda\cap\mathfrak a^*=\Gamma\cdot\lambda$$ hold for every $\lambda\in\mathfrak a^*$?

**Motivation**: If true, it would imply that the all invariants
of $V\oplus V^*$ as a representation of $\mathbb C^\times\times E_6$ would be pull-backs of coadjoint invariants under the moment map.

**Some details**: The roots of $E_6$ are (we use Bourbaki conventions)
$$ \pm\epsilon_i\pm\epsilon_j\quad1\leq i<j\leq5\quad\mbox{(for $Spin(10)$)} $$
and
$$ \frac12(\epsilon_8-\epsilon_7-\epsilon_6+\sum_{i=1}^5\nu_i\epsilon_i)\quad\Pi_{i=1}^5\nu_i=1\quad\mbox{(for the half-spin representation)}. $$
The $27$-dimensional irreducible representation $V$ of $E_6$ and its contragredient $V^*$ have respectively highest weights
$$\lambda_1 = \frac23\Omega\quad\mbox{and}\quad\lambda_6=\frac13\Omega+\epsilon_5 $$
where $\Omega=\epsilon_8-\epsilon_7-\epsilon_6$. These are minuscule weights, which means that the Weyl group $W$ acts transitively on the weights of $V$ (resp. $V^*$); in particular they all have multiplicity one. The weights of $V$ can be listed as
$$ \frac23\Omega,\ \frac16\Omega-\frac12\sum_{i=1}^5\nu_i\epsilon_i\ (\Pi_{i=1}^5\nu_i=1), -\frac13\Omega\pm\epsilon_i\ (1\leq i\leq5).$$
In particular,
$$ \lambda_1,\ -\lambda_6,\ -\lambda_1+\lambda_6 $$
are weights of $V$; the longest element in $W$ maps the highest weight $\lambda_1$ and the lowest weight $-\lambda_6$ one to the other and thus fixes $-\lambda_1+\lambda_6$. One can indeed find elements in $W$ permuting
$\lambda_1$, $-\lambda_6$, $-\lambda_1+\lambda_6$.
The angle between $\lambda_1$ and $\lambda_6$ is $60^o$. Now it is not difficult to see that $\Gamma$ is the Coxeter group of type $A_2$.

**Edit**: Here is a naive argument which appears to give an answer.
Suppose $w\xi_1=\xi_2$ where $w\in W$ and $\xi_1$, $\xi_2\in\mathfrak a^*$.
Since the action of $W$ preserves the real span $\mathfrak a^*_{\mathbb R}$ of $\lambda_1$, $\lambda_6$, we may assume $\xi_1$, $\xi_2\in\mathfrak a^*_{\mathbb R}$.
Since $\Gamma$ is a subquotient of $W$ and the action of $\Gamma$ on $\mathfrak a^*$ is generated by reflections on the real lines through $\lambda_1$, $\lambda_6$,
we may replace $\xi_1$ and $\xi_2$ by suitable $\Gamma$-conjugates and assume both are linear combinations of $\lambda_1$, $\lambda_6$ with non-negative coefficients. Now $\lambda_1$, $\lambda_6$ are fundamental weights of $E_6$, so
this implies that $\xi_1$, $\xi_2$ belong to the positive Weyl chamber of $E_6$ and thus $w=1$.

**Update**: The argument in the first edit seems OK, but about the above motivation, it is *not* true that it implies that the invariants are all pull-backs of coadjoint invariants.

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