Let $p$ be a prime number and $\mathbb{F}_{p}$ be a finite field of order $p$. Let $G=GL_{n}(\mathbb{F}_{p})$ denote the general linear group and $U_{n}$ denote the unitriangular group of $n\times n$ upper triangular matrices with ones on the diagonal, over the finite field $\mathbb{F}_{p}$. Let $T$ be the diagonal subgroup of $G$.
Definition: Let $D\in T$. A diagonal automorphism $\varphi _{D}$ of $U_{n}$ is an automorphism defined by $\varphi_{D}(M)=DMD^{-1}$. A subgroup U of $U_{n}$ is called invariant under the diagonal automorphisms of $U_{n}$ if $\varphi_{D}(M)\in U$ for all $M\in U$.
Question: Do the subgroups of order $p^{2}$ of $U_{n}$ are invariants under the diagonal automorphisms of $U_{n}$.
There are in general some subgroups of order $p^2$ which are not $T$-invariant when $p$ is odd.
For instance, take $$ H:=\left\{ \begin{pmatrix} 1 & a & 0 & b\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & a\\ 0 & 0 & 0 & 1 \end{pmatrix} \mid a,b\in\mathbb{F}_p \right\}, $$ It is not conjugated to itself by $\text{diag} (2,1,1,1)$.
Added later
It occurred to me that it might be interesting to know whether the normal subgroups of the unitriangular group are fixed by the diagonal group. The facts of the matter are (when $n\geqslant 3$ and $p$ is odd):
there are exactly $p+1$ such subgroups, namely the maximal subgroups of the second centre;
these may labelled in a natural way by the ratios $\alpha : \beta$ for all $\alpha,\beta\in\mathbb{F}_p$;
of these, the subgroups labelled $0$ and $\infty$ are fixed by the diagonal group, and the others permuted transitively.
The calculations which establish these facts are fairly routine; one begins by establishing that the centre consists of those unitriangular matrices whose only non-zero off-diagonal entry is the last in the first row, and that the second centre consists of those unitriangular matrices whose only non-zero off-diagonal entries are the last two in the first row and the first two in the last column.