Let $p$ be a prime number and $G=GL_n ( \mathbb{Z} / p \mathbb{Z} )$. Consider the set $U$ of uppertriangular matrices of $G$ having entries of $1$ on the diagonal. The cardinality of $U$ is $p^{\frac {n(n1)} 2}$ and $U$ is a subgroup of $G$, in particular $U$ is a Sylow $p$subgroup of $G$. It is well known that the Sylow $p$subgroups of a group $G$ are conjugate, and every $p$subgroup $H$ of $G$ is contained in some Sylow $p$subgroup of $G$. Then there exists $g\in G$ such that $H\leq gUg^{1}$, which allows us to compute the number of conjugacy classes of elementary abelian subgroups of rank $2$ ($H=( \mathbb{Z} / p \mathbb{Z} ) ^2$) in the Sylow $p$subgroup $U$. Any help would be appreciated so much. Thank you all.
I would say the problem is open.
Consider the case where $p>n$, i.e. every element has order $p$ in $U$.
The number of such subgroups is $(U·k(U)1\frac{U1}{p1}(p^21))\frac{1}{(p^21)(p^2p)}$, where $k(U)$ is the number of conjugacy classes of $U$.
Let me explain the formula. The number of commuting pairs in any finite group G is $G⋅k(G)$. Such pairs may
Generate a trivial group (only 1 choice)
Generate a group of order $p$. There're $\frac{U1}{p1}$ groups of order $p$ in $U$, and for each group, there're $p^21$ choices to generate the group with $2$ elements.
Generate a group of order $p^2$. The group is surely rank $2$, and has $(p^21)(p^2p)$ generating pairs. Dividing the number of such pairs in $U$ by $(p^21)(p^2p)$ gives the answer.
So, the question is equivalent with determining the number of conjugacy classes of $U$, which is the Higman Conjecture.

$\begingroup$ Thank you, sir, but the subgroup U defined in my question is not as in Higman Conjecture. I think there is some confusion. $\endgroup$ – Nourddine Snanou Oct 6 '19 at 18:42

1$\begingroup$ What's the difference? Aren't they both Sylow psubgroups? $\endgroup$ – LeechLattice Oct 7 '19 at 3:41

$\begingroup$ So I am confused because I think that the number of conjugacy classes of $U$ is the index of its normalizer in $G=GL_n (\mathbb{Z} / p \mathbb{Z}) $ ($k(U)=[G,N_{G}(U)]$). $\endgroup$ – Nourddine Snanou Oct 7 '19 at 9:57