Number of distinct conjugacy classes of the multiplicative group of $3\times 3$ upper-triangular matrices with all diagonal entries $1$ Let $F$ be a finite field of order $q$ ; let $U(3,F)$ be the multiplicative group of $3\times 3$ upper-triangular matrices with entries from $F$ and with all diagonal entries $1$ . Then is it true that the number of distinct conjugacy classes of  $U(3,F)$ is a polynomial in $q$ with integer co-efficients ?
 A: This group has an interesting property (second, below): denote group by $G$. Then
(1) $|G|=q^3$ and $|Z(G)|=|G'|=q$; center gives obviously $q$ number of  singleton classes.
(2) Outside center, any two elements have same conjugacy class size and is equal to $|G'|=|Z(G)|$.
(3) With the help of (1) and (2), you can easily determine number of conjugacy classes in $G$, which turns out to be very specific polynomial in $q$; I am saying here very specific because in higher orders of matrices, the conjugacy class sizes are be more than two (i.e. property in (2) is not true for higher rank matrix groups)  and so polynomial becomes more complicated.

For (2), do following: show that center of this group consists of matrices in group with off-diagonal zero and top-right corner arbitrary (i.e. $a_{12}=a_{23}=0; a_{13}$ arbitrary. 
Then take arbitrary element 
$$A=\begin{bmatrix} 1 & a & b\\  & 1 & c \\  & & 1\end{bmatrix}, \,\,\,\, a\neq 0 \mbox{ or } b\neq 0.$$
Try to find order of centralizer this element in $G$, and to do simplifications, we can assume that $c=0$ because we can mod out this element by $Z(G)$ (centralizer of $t$ and centralizer of $tz$ are equal for any $z\in Z(G)$).
You will see that centralizer has order $q^2$, hence number of conjugates is $q$ which is also $|G'|$.
