It is my understanding that SU(3) acts on $\mathbb{C}^3$ the same way SO(3) acts on $\mathbb{R}^3$, i.e. as proper rotations.
If this is the case, then the orbit of $(1,0,0) \in \mathbb{C}^3$ should be the unit sphere, i.e. vectors in $\mathbb{C}^3$ with norm $1$. However this is wrong according to my teacher who says $M$, defined as $(z_1, z_2,z_3)$ with $\sum z_i^2 =1$ is not an orbit of SU(3).
The orbit is definitely contained in unit normed vectors as $v^\dagger v = v^\dagger U^\dagger U v = (Uv)^\dagger (Uv)$. However, I'm not sure what I am doing wrong.
Can someone please help see why $(1,0,0)$ cannot be mapped to all vectors of the $(z_1, z_2,z_3)$ with $\sum z_i^2 =1$ using $SU(3)$. Also, what is the isotropy group in this case? I would think it's the SU(2) subgroup rotating orthogonally to $(1,0,0)$, i.e. matrices of the form
$$ C=\left[ \begin{array}{cc} 1 & 0 \\ \hline 0 & U \end{array}\right] $$
where U is in SU(2).
My guess is that somehow, we escape the subspace $M$, because she then asks about SO(3)'s action on it. With SO(3), we can think of a matrix acting on the real vector sum $(\bar x+i \bar y)$, and thus the norm of $\bar x$ and $\bar y$ are preserved, which leads to the complex norm being preserved, i.e. $\sum z_i^2 =1$.