Fundamental group of nilpotent matrices $G=SL_n(\Bbb C)$ acts on $N=\{A \in M_n(\Bbb C)| A^n=0 \}$ by conjugation and $X=$the orbit of $A_0=\begin{bmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 &0& 0 & \dots & 1 \\ 0 &0& 0 & \dots & 0 \end{bmatrix}$ is dense in $N$. How to compute the fundamental group of $X$ or equivalently $N$? By the long exact sequence of a fiber bundle I know $ \pi_1(X)$ has order $n$, is it $\Bbb Z/n \Bbb Z$?
 A: I think you've probably got this yourself by this point but $\pi_1(X)$ is indeed $\mathbb{Z}_n$ and to see this we do in fact use the long exact homotopy sequence. The conjugation action of $SL_n(\mathbb{C})$ on the orbit $X=SL_n(\mathbb{C})\cdot A_0$ is transitive by definition so produces a principal fibring 
$G\rightarrow SL_n(\mathbb{C})\xrightarrow{ev} X$
where the subgroup $G\leq SL_n(\mathbb{C})$ is the stabiliser of $A_0$. Thus $g\in SL_n(\mathbb{C})$ lies in $G$ if and only if it satisfies $g\cdot A_0 =A_0 \cdot g $. 
Working loosely, $A_0$ has components $(A_0)_{ij}=\delta_{i,j-1}$, so the previous equation leads to the condition $g_{i,j-1}=g_{i+1,j}$ for the components of $g\in G$ (this is sloppy since you still have to work to sort out the boundary cases, sorry). The end result is that the elements $g\in G$ have the following form
$g=\begin{bmatrix} a_1 & a_2 & a_3 & \dots & a_n \\ 0 & a_1 & a_2 & \dots & a_{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 &0& 0 & \dots & a_{2} \\ 0 &0& 0 & \dots & a_1 \end{bmatrix}$
with zeros below the diagonal, and constant complex values along the diagonal and each subdiagonal above it. Since $G\leq SL_n(\mathbb{C})$ we also have the condition of unit determinant, so $det(g)=(a_1)^n=1$ tells us that the constant value of $a_1$ along the main diagonal lies in the complex $n^{th}$roots of unity $\mathbb{Z}_n\leq \mathbb{C}$.
Now in fact $G\simeq \mathbb{Z}_n$, since we have the homotopy $F((a_1,\dots,a_n),t)=(a_1,(1-t)a_2,\dots,(1-t)a_n)$ where we write $g=(a_1,\dots,a_n)$ for $g\in G$ as above. This homotopy $F$ starts at the identity of $G$ and contracts down to the diagonal subgroup $\mathbb{Z}_n\leq G$. In particular $\pi_0(G)=\mathbb{Z}_n$ and $\pi_1(G)=0$. Since $SL_n(\mathbb{C})$ is simply connected we also have that $\pi_0(SL_n(\mathbb{C}))=0$ and $\pi_1(SL_n(\mathbb{C}))=0$, so the homotopy exact sequence of the fibration introduced above gives an isomorphism 
$\pi_1(X)\cong \pi_0(G)\cong\mathbb{Z}_n$
For its higher homotopy groups we have $\pi_n(X)\cong\pi_n(SL_n(\mathbb{C}))\cong\pi_n(SU(n))$.
