The orbit of a regular nilpotent matrix is connected iff n is odd

Idea. The class $\mathcal O_N$ is the image of $GL_n(\mathbb{R})$ under $P\mapsto PNP^{-1}$. It is connected as soon as $P$ and some $P'$ of the opposite determinant sign give the same matrix. This happens for odd $n$ ($P'=-P$), and fails for even $n$.

Step 0: description of the orbit. $\mathcal O_N=\{A:A^n=0,\ A^{n-1}\ne0\}$. Indeed, such an $A$ has a vector $u$ with $A^{n-1}u\ne0$; then $u,Au,\dots,A^{n-1}u$ are linearly independent (if $\sum c_kA^ku=0$ with $c_j$ the first nonzero coefficient, apply $A^{n-1-j}$ to get $c_jA^{n-1}u=0$), so they form a basis in which $A$ acts as $N$. With $P=[u,Au,\dots,A^{n-1}u]$ we get $A=PNP^{-1}$.

Step 1: $n$ odd. For $P\in GL_n^-$, the matrix $-P$ has $\det(-P)=(-1)^n\det P>0$ and $(-P)N(-P)^{-1}=PNP^{-1}$. So $\mathcal O_N=\{PNP^{-1}:P\in GL_n^+\}$, the continuous image of the connected set $GL_n^+$. Hence $\mathcal O_N$ is connected.

Step 2: $n$ even the invariant. Call $v\in\mathbb{R}^n$ *cyclic for $A$* if $A^{n-1}v\ne0$; by Step 0, then $K_v(A)=[v,Av,\dots,A^{n-1}v]$ is invertible. Put $\varepsilon(A)=\operatorname{sign}\det K_v(A)$.

*Independence of $v$.* Let $v,v'$ be cyclic for $A$. Since $K_v(A)$ is a basis, $v'=q(A)v$ for a polynomial $q(t)=c_0+c_1t+\dots+c_{n-1}t^{n-1}$. Then $A^kv'=q(A)A^kv$, so $K_{v'}(A)=q(A)K_v(A)$ and
$$\det K_{v'}(A)=\det q(A)\cdot\det K_v(A).$$
Because $A$ is nilpotent, $q(A)=c_0I+(\text{nilpotent})$, which has the single eigenvalue $c_0$, so $\det q(A)=c_0^n$. As $q(A)$ is invertible ($K_{v'}$ is), $c_0\ne0$, and since $n$ is even $c_0^n>0$. So the sign is the same for $v$ and $v'$. (For odd $n$ this argument breaks down: $c_0^n$ can be negative.)

Step 3: $\varepsilon$ is continuous on $\mathcal O_N$. Let $A_0\in\mathcal O_N$ and let $v$ be cyclic for $A_0$. The entries of $K_v(A)$ are polynomial in $A$, so $\det K_v(A)\ne0$, with the same sign as at $A_0$, for all $A$ in a neighbourhood $U$ of $A_0$. For $A\in U\cap\mathcal O_N$ this says that $v$ is cyclic for $A$ and $\varepsilon(A)=\operatorname{sign}\det K_v(A)=\varepsilon(A_0)$. So $\varepsilon$ is locally constant.

Step 4: the two components. For $A=PNP^{-1}$ take $v=Pe_1$: then $A^kv=PN^ke_1=Pe_{k+1}$ and $K_v(A)=P$. So $\varepsilon(PNP^{-1})=\operatorname{sign}\det P$, and
$$\mathcal O^\pm=\{PNP^{-1}:\pm\det P>0\}=\varepsilon^{-1}(\pm1).$$
Both are nonempty, each is the continuous image of the connected set $GL_n^\pm$, hence connected, and they are open and closed in $\mathcal O_N$ because $\varepsilon$ is continuous. So $\mathcal O_N$ has exactly two components.

Check. $n=2$: $\mathcal O_N$ is the set of nonzero nilpotent matrices of $M_2(\mathbb{R})$, which indeed has two components (the two nappes of the cone $\{A^2=0\}\setminus\{0\}$). $n=1$: $\mathcal O_N=\{0\}$, connected. A numerical experiment with random $P$ ($n=4,6$) shows $\operatorname{sign}\det K_v(PNP^{-1})\cdot\operatorname{sign}\det P=+1$ for random cyclic $v$; for $n=3,5$ both signs occur.

Remark. In group-theoretic terms, $\mathcal O_N\cong GL_n/C(N)$ where the centraliser is $C(N)=\{a_0I+a_1N+\dots+a_{n-1}N^{n-1}:a_0\ne0\}$ with $\det=a_0^n$. If $n$ is odd, $C(N)$ meets $GL_n^-$ (take $-I$) and the quotient is connected; if $n$ is even $C(N)\subset GL_n^+$ and the orbit inherits two components.

Lesson. Whether a conjugacy orbit is connected is decided by whether the centraliser meets the non-identity component of $GL_n(\mathbb{R})$.