Order of normalizer of upper triangular matrices subgroup in $GL_n(\mathbb{F}_p)$ Let $p$ a prime and $\mathbb{F}_p$ the finite field of $p$ elements. For $n \geq 2$ and $G=GL_n(\mathbb{F}_p)$, let $T\subseteq_{sg}G$ the subgroup of upper triangular matrices with all entries in principal diagonal equal to $1$. Show that $|N_{G}(T)|=(p-1)^{n}p^{n(n-1)/2}$.
Using a combinatorial argument I already showed that $|T|=p^{n(n-1)/2}$, but I do not know how to find the order of the normalizer and why it is the order of $T$ multiplied by $(p-1)^n$... 
Any help would be appreciated.
Thanks.
 A: The key to understanding triangular matrices conceptually is relating them to flags.  Specifically, let $V_k\subseteq \mathbb{F}_p^n$ be the subspace spanned by the first $k$ standard basis vectors.  Note that an invertible matrix $A$ is upper triangular iff $A(V_k)=V_k$ for all $k$.  More specifically, a matrix $A$ is in $T$ iff $A-I$ is nilpotent and $A(V_k)=V_k$ for all $k$.  Conversely, we can define the $V_k$ from $T$: $V_k$ is the set of vectors that are annihilated by $(A-I)^k$ for all $A\in T$.
This makes it easy to understand conjugates of $T$.  In particular, since we can define the $V_k$ in terms of $T$ and vice versa, for any $B\in G$ we have $BTB^{-1}=T$ iff $B(V_k)=V_k$ for all $k$.  Explicitly, $BTB^{-1}$ is exactly the set of $A$ such that $A-I$ is nilpotent and $A(B(V_k))= B(V_k)$ for all $k$, and conversely $B(V_k)$ is the set of vectors annihilated by $(A-I)^k$ for all $A\in BTB^{-1}$.  More generally, the conjugates of $T$ are in bijection with sequences of subspaces $W_0\subset W_1\subset\dots\subset W_n$ with $\dim W_k=k$ for each $k$, with such a sequence corresponding to $BTB^{-1}$ for any $B$ such that $B(V_k)=W_k$ for each $k$.  (Such sequences are called complete flags in $\mathbb{F}_p^n$.)
So, $N_G(T)$ is exactly the set of $B\in G$ such that $B(V_k)=V_k$ for all $k$.  That is, it is exactly the set of invertible upper triangular matrices.  I will leave it to you to finish the problem by counting how many such matrices there are.
