Letting $SL_n(\mathbb{C})$ act on $\mathbb{C}^n$, what orbits do you get?
Answer: Clearly $A0 = 0$ for all $A \in SL_n(\mathbb{C})$. Also as $A$ is invertible, $A0 = 0 \iff v = 0$, so $0$ forms a singleton orbit. Now given any two vectors $ v \neq w \in \mathbb{C}^n$, there is a matrix $A \in SL_n(\mathbb{C})$ such that $Av = w$, so the orbits are $0$ and $\mathbb{C}^n \setminus 0$.
Is this last assertion about there existing an A true, and if so, why (using elementary group theory)
Let $v=v_1$ be any nonzero vector in $\mathbb{C}^n \setminus \{0\}$. Complete it to a basis $v_1, v_2, \dots, v_n$ of $\mathbb{C}^n$ and let $A$ be the matrix whose columns are the $v_i$. Then $A$ is invertible and satisfies $Ae_1 = v$. Assuming $n \geq 2$, change the matrix $A$ by replacing any of the $v_i$ with $i \geq 2$ by $((\det{A})^{-1})v_i$. Call the resulting matrix $A'$. Then $A'$ is in $\text{SL}_n{\mathbb{C}}$ and satisfies $A'e_1 = v$. This shows that the induced action of $\text{SL}_n(\mathbb{C})$ on $\mathbb{C}^n \setminus \{0\}$ is transitive (since we have shown that the orbit of $e_1$ is $\mathbb{C}^n \setminus \{0\}$).
Note that the assertion is not quite true for $n=1$.
For $n > 1$, though, it is a matter of showing that if you fix the first column of a matrix $A$ to be an arbitrary non-zero vector, then you can complete $A$ to an element of the special linear group. Without lack of generality, suppose the first column is $$ \begin{bmatrix} a_{1}\\a_{2}\\ \vdots \\a_{n} \end{bmatrix}, $$ with $a_{1} \ne 0$. Then take $$ A = \begin{bmatrix} a_{1} & 0 & 0 & \dots & 0\\ a_{2} & a_{1}^{-1} & 0 & \dots & 0\\ a_{3} & 0 & 1& \dots & 0\\ &&& \ddots \\a_{n} & 0 & 0 & \dots & 1\\ \end{bmatrix} $$