In an assignment, I am asked to find the number of nilpotent $4 \times 4$-matrices over af finite field $\mathbb{F}_p$, $p$ prime. In the process I shall find the size of the stabilizer (General Linear group acts by conjugation) of the matrix $$A=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ I believe that my answer is incorrect, and hope someone can give a hint or present a solution.
Attempted solution: I write the equation $xA=Ax$, $x=(x_{ij})$, and find that $$ x=\begin{pmatrix} x_{11} & x_{12} & x_{13} & x_{14} \\ 0 & x_{11} & 0 & x_{13} \\ x_{31} & x_{32} & x_{33} & x_{34} \\ 0 & x_{31} & 0 & x_{33} \end{pmatrix} $$ Requiring that $x$ be invertible, I come to the conclusion, that the number of such matrices $x$ is $$p^5(p-1)^2+p^4(p-1)^3+p^5(p-1)^2+p^4(p-1)^3 = 2p^4(p-1)^2(2p-1),$$ which is then the size of the stabilizer.
I believe that the answer is incorrect, and need clearification.
We have $x$ singular if and only if $\det x=0$, and since $$\det x = x_{11}(x_{11}x_{33}^2-x_{31}x_{33}x_{13})+x_{31}(x_{13}^2x_{31}-x_{11}x_{33}x_{14}) = (x_{11}x_{33}-x_{13}x_{31})^2$$ we can divide into cases and find that there are $p(p^2+p-1)$ possibilities such that $\det x=0$, giving $p^4-p(p^2+p-1)$ possibilities such that $\det x \neq 0$, and in total $$p^4(p^4-p(p^2+p-1))=p^5(p+1)(p-1)^2$$ possibilities for $x$ to be invertible - so this is also the size of the stabilizer.