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The classical problem regarding the action of symplectic group on its Lie algebra gives rise to the following question in the finite field case.

Let $\mathbb F_p$ be a finite field. Then the symplectic group over $\mathbb F_p$ acts by conjugation on the set of matrices over $\mathbb F_p$ that satisfy $\Omega A + A^t \Omega = 0$, $\Omega$ is the skew symmetric matrix

$$ \begin{pmatrix} 0 & I \\\\ -I & 0 \end{pmatrix} $$

where $I$ is identity matrix. What are the orbits of this action?

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This problem is answered in a paper by Burgoyne and Cushman. I don't have the reference to hand.

This also came up in Classification of adjoint orbits for orthogonal and symplectic Lie algebras?

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The subspace in question is $\mathfrak{sp}_{2n}\textbf{F}_p$; explicitly, it consists of matrices of the form $\begin{pmatrix}A&B\\C&-A^\top\end{pmatrix}$ where $B=B^{\top}$ and $C=C^{\top}$.

Although it won't tell you everything about the orbits, the description of $\mathfrak{sp}_{2n}\textbf{F}_p$ as an $\text{Sp}_{2n}\textbf{F}_p$-representation is known. This can be found in Hogeweij, "Almost-classical Lie algebras. I." Nederl. Akad. Wetensch. Indag. Math. 44 (1982), no. 4, 441-452, but it is hard to extract the answer from that paper, so I'll briefly give the argument. [Edit: I am no longer sure how to extract the argument below from that paper, but I believe everything below is self-contained and can be directly verified except the assertions that various representations are irreducible.]

If $p$ is odd, then $\mathfrak{sp}_{2n}\textbf{F}_p$ is irreducible with highest weight $2\omega_1$, where $\omega_1,\ldots,\omega_n$ are the fundamental weights for $\text{Sp}_{2n}\textbf{F}_p$.

If $p=2$, we proceed as follows. Let $H\approx \textbf{F}_2^{2n}$ be the standard representation of $\text{Sp}_{2n}\textbf{F}_2$, so the explicit description of $\mathfrak{sp}_{2n}\textbf{F}_2$ above identifies it with a subspace of $\mathfrak{gl}_{2n}\textbf{F}_2\approx H^*\otimes H$. The symplectic form gives an isomorphism $H^*\cong H$ so we can identify $\mathfrak{sp}_{2n}\textbf{F}_2$ with a subspace of $H\otimes H$. The corresponding subspace is precisely the invariant subspace $\Gamma^2 H=(H\otimes H)^{\mathbb{Z}/2}$; so in particular the original question is asking about the orbits of $\text{Sp}_{2n}\textbf{F}_2$ on $\mathfrak{sp}_{2n}\textbf{F}_2\cong\Gamma^2 H$. [I am grateful to Andy Putman for pointing out a mistake in the earlier version of this answer, where I mistakenly had $\text{Sym}^2 H$ instead of $\Gamma^2 H$.]

First, let us discuss the structure of $\Gamma^2 V$ as a $\text{GL}(V)$ representation in characteristic 2, before specializing to $\text{Sp}_{2n}\textbf{F}_2$. In characteristic 2 we have short exact sequences of $\text{GL}(V)$-representations: $$0\to \textstyle \bigwedge^2 V\to \bigotimes^2 V\to \text{Sym}^2 V\to 0$$ $$\textstyle 0\to \Gamma^2 V\to \bigotimes^2 V\to \bigwedge^2 V\to 0$$ and also $$\textstyle0\to V(1)\to \text{Sym}^2 V\to \bigwedge^2 V\to 0$$ $$\textstyle0\to \bigwedge^2 V\to \Gamma^2 V\to V(1)\to 0$$ (Our interest will be in the last one.) Note that here $V(1)$ denotes the "Frobenius twist" of $V$, which is the same vector space as $V$ but with the $\text{GL}(V)$ action twisted by Frobenius (so if $V$ is a vector space over $\mathbf{F}_2$ itself, there is no difference). The embedding $V(1)\to \text{Sym}^2 V$ sends $x\mapsto x\cdot x$, which is linear since $(x+y)^2 = x^2+y^2$ (but note that it does not commute with the action of e.g. scalar matrices in $\text{GL}(V)$, which is why the domain must be $V(1)$ instead of just $V$).

Returning to $\text{Sp}_{2n}\textbf{F}_2$, we have an exact sequence $0\to \bigwedge^2 H\to \Gamma^2 H\to H\to 0$. Note that in terms of our description $\mathfrak{sp}_{2n}\textbf{F}_2=\begin{pmatrix}A&B=B^{\top}\\C=C^{\top}&-A^\top\end{pmatrix}$, the subspace $\bigwedge^2 H$ corresponds to those matrices for which both $B$ and $C$ have all diagonal entries equal to 0. So we need to consider $\bigwedge^2 H$ as an $\mathfrak{sp}_{2n}\textbf{F}_2$-representation. This has two invariant subrepresentations.

First, there is a trivial representation spanned by the vector $\omega=a_1\wedge b_1+\cdots+a_n\wedge b_n$ (in our matrix representation, this is the matrix $\begin{pmatrix}I&0\\0&I\end{pmatrix}$).

Second, there is the kernel $K$ of the contraction $c\colon \bigwedge^2 H\to \textbf{F}_2$, defined by $a_i\wedge b_i\mapsto 1$, $a_i\wedge a_j\mapsto 0$, $b_i\wedge b_j\mapsto 0$, and $a_i\wedge b_j\mapsto 0$. (In terms of matrices, this contraction sends $\begin{pmatrix}A&B\\C&-A^\top\end{pmatrix}\mapsto \text{tr} A$, so $K$ is the subspace where $A$ has even trace and $B$ and $C$ have no diagonal entries.)

Note that under the contraction $c$ the invariant element $\omega$ is taken to $n\in \textbf{F}_2$. Therefore when $n$ is odd the element $\omega$ provides a section of $c$, giving a direct sum decomposition $\bigwedge^2 H\cong \textbf{F}_2\oplus K$. In this case $K$ is irreducible.

When $n$ is even, on the other hand, we have $\langle\omega\rangle\subset K\subset \bigwedge^2 H$; in this case $K/\langle\omega\rangle$ is irreducible.

If I'm not mistaken, the invariant subrepresentations of $\mathfrak{sp}_{2n}\textbf{F}_2$ are (regardless of the parity of $n$) $\langle\omega\rangle$, $K$, and $\bigwedge^2 H$. (There could possibly be some that project onto $H$, but I believe/suspect any such subrepresentation must be everything.)

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