Unitary orbit of the Jordan matrices Let
$$
\mathcal{J}=\{A\in M_n(\mathbb{C}):\ A \text{ is a Jordan matrix}\}
$$
Then it is well-known that the similary orbit of $\mathcal{J}$ is all of $M_n(\mathbb{C})$.

What is the unitary orbit of $\mathcal{J}$? Is it dense?

It cannot be all of $M_n(\mathbb{C})$, because every matrix in $\mathcal{J}$ and its unitary conjugates have the property that eigenvectors corresponding to different eigenvalues are orthogonal to each other.
 A: It cannot be dense except in the trivial case of $n=1$.  The (real) dimension of $U(n)$ is $n^2$ while the dimension of $Gl_n(\mathbb{C})$ is $2n^2$  Since $\mathcal{J}$ has dimension $n$ (in the sense that it's a union of the diagonal matrices together with unions of various Jordan blocks things of smaller dimension since we get to choose fewer eigenvalues), the $U(n)$ orbit through $\mathcal{J}$ has dimension at most $n^2+n$.
But $n^2 + n\leq 2n^2$ unless $n=n^2$, i.e., $n=0$ or $n=1$.
A: It is the set of $n \times n$ matrices $A$ such that there is an orthormal basis $\{u_j:\ j=1\ldots n\}$ that is the union of blocks $\{u_j:\ a \le j \le b\}$ where $u_a$ is an eigenvector of $A$ for some eigenvalue $\lambda$ and $A u_j = \lambda u_j + u_{j-1}$ for 
$a+1 \le j \le b$.  Well, that's just restating the fact that the matrix for $A$ in this basis should be in Jordan form.
EDIT: here's a necessary condition, I'm not sure it's sufficient.  $A = B + N$ where $B$ is a normal matrix and $N$ a nilpotent matrix that commutes with $B$, and such that
$N^k (N^k)^*$ and $(N^k)^* N^k$ are orthogonal projections for each positive integer $k$.
