Problem: $V$ is linear space with dimension $n$, $G$ is a group of all invertible transformation on $V$ that its representation matrix is an upper triangular matrix. Considering $G$ acts on $V$ by $gx:=g(x)$, compute all its orbits.
I know an orbit: $\{0\}$ and if the $k$th coordinate component $x_k\neq 0$ while $x_l=0$ for all $l＞k$, I think $\mathrm{Orb}(x)$ is the family of all elements satisfied the $k$th $\neq 0$ while $l$th$=0$ for every $l＞k$. However, I can't achieve this idea.
$G$ consists of all $n \times n$ invertible upper triangular matrices, and $G$ acts (from the left) on vector space $V$.
The claim is that the orbits of $G$ on $V$ are the sets $O_k$ with $0 \le k \le n$, where $O_0 = \{0\}$, and for $k > 0$, $O_k$ consists of those vectors $v$ with $v_k \ne 0$ and $v_j = 0$ for all $j>k$.
Since the matrices in $G$ are invertible, they have nonzero elements on the diagonal, and so, for each $k$, $G$ maps vectors in $O_k$ to vectors in $O_k$.
It remains to prove that each $O_k$ is a complete orbit. For this, fix $k$, let $x$ be the vector with $x_k=1$ and $x_j = 0$ for $j \ne k$, and let $y$ be an arbitrary vector in $O_k$.
It is sufficient to prove that there is some element of $G$ that maps $x$ to $y$. But we can just choose any matrix having $y$ as its $k$-th column - the other entries do not matter - and that will map $x$ to $y$.