Conjugacy of p-subgroups in $GL_{5}(\mathbb{F}_{p})$?. Let $U_{5}$ denote the unitriangular group of $5\times 5$ upper triangular matrices with ones on the diagonal, over the finite field $\mathbb{F}_{p}$. Let $H=\left. \left\{
A=\begin{pmatrix}
1 & 0 & 0 & a &d \\
0 & 1 & 0 & b &e \\
0 & 0 & 1 & c &f \\
0 & 0 & 0 & 1 &0 \\
0 & 0 & 0 & 0 &1%
\end{pmatrix}%
\right| a,b,c,d,e,f \in \mathbb{F}_{p}\right\}$.
and
$K=\left. \left\{B=
\begin{pmatrix}
1 & 0 & a' & b' &c' \\
0 & 1 & d' & e' &f' \\
0 & 0 & 1 & 0 &0 \\
0 & 0 & 0 & 1 &0 \\
0 & 0 & 0 & 0 &1%
\end{pmatrix}%
\right| a',b',c',d',e',f' \in \mathbb{F}_{p}\right\}$ be two subgroups of $GL_{5}(\mathbb{F}_{p})$. The subgroups $H$ and $K$ are maximal abelian normal in $U_{5}$ (See for example Exercise $3$ p. $94$ of the Book {M. Suzuki, Group theory I}).
Does the subgroups $H$ and $K$ conjugate in $GL_{5}(\mathbb{F}_{p})$?.
I think the answer is No but I don't sure what to do about it.
My try to this question:
Let $V$ be a vector of $\mathbb{F}_{p}^{5}$. H and K are not conjugate since
$I(\mathbb{F}_{p}[H])V$ is a 3-dimensional vector space but $I(\mathbb{F}_{p}[K])V$ is just a 2-dimensional. Here, $I$ denotes the augmentation ideal.
Could anyone please tell me if my try is correct or provide a defferent approche?
Thank you in advance.

Prove that the fixed point space of $H$ on $\mathbb{F}_p^5$ has dimension $3$.
Prove that the fixed point space of $K$ on $\mathbb{F}_p^5$ has dimension $2$.
Conclude that $H$ and $K$ cannot be conjugate in $GL_5(\mathbb{F}_p)$.
