I don't know the answer, but I am posting to make sure I understand the question and give some partial ideas.

As Ben says, the question is hard to follow. I am interpreting it this way:

Consider the cone of nilpotent elements in $\mathfrak{so}_{2n}$. How does it break into $SO_{2n}$ orbits under the adjoint action? How is this related to the decomposition according to Jordan normal form? (Note: the decomposition according to Jordan normal form can be thought of as the $SL_{2n}$ orbits for the nilpotent matrices in $\mathfrak{sl}_{2n}$.)

Some partial thoughts. The Lie algebra $\mathfrak{so}_{2n}$ is skew symmetric matrices. Any skewsymmetric matrix has even rank. Moreover, if $A$ is skewsymmetric, so is every odd power of $A$. So $A$, $A^3$, $A^5$ etcetera all have even rank. This implies that, if $\lambda$ is the partition of $2n$ coming from the Jordan form of $A$, then every even part of $\lambda$ occurs with even multiplicity. So, when $2n=4$, the only partitions that occur are $31$, $22$ and $1111$. So we have determined which Jordan forms appear.

In the particular case $n=4$, we have $\mathfrak{so}_4 \cong \mathfrak{sl}_2 \oplus \mathfrak{sl}_2$. Using this isomorphsim, one checks that $n_1 \oplus n_2$ is nilpotent if and only if $n_1$ and $n_2$ are. We get Jordan form $31$ if both $n_1$ and $n_2$ are nonzero nilpotents; we get $22$ is one is zero and the other isn't; and we get $1111$ if both are nilpotent.

So we see that Jordan form $22$ splits into two orbits, according to which of $n_1$ and $n_2$ is nonzero. Presumably, rajamanikkam knows some theorem which says that this happens whenever all the parts of $\lambda$ are even, and would like to know how to make the theorem explicit.

In general, I know how I would attack this problem. The Jordan form of $A$ fixes the
Jordan form of $S_{\mu}(A)$, for any Schur functor $S_{\mu}$. But the $SO_{2n}$ conjugacy class fixes the Jordan form $A$ acting on any $\mathfrak{so}_{2n}$ representation. In particular, the Jordan form of the action of $A$ on the spin representations should give additional data, allowing us to separate distinct $SO_{2n}$ orbits.

I don't know the details of how this works, but I'm sure someone reading this does!