As Jim Belk remarks in his answer, the map $\sigma:z\in\mathbb{C}^n\to\sqrt{-1}zz^\ast\in\mathfrak{u}(n)$ is smooth and $(\mathbb{R}_+\times U(n))$-equivariant. Here the actions are given by $(t,C).z=tCz$ over $\mathbb{C}^n$, and $(t,C).A=t^2 CAC^\ast$ over $\mathfrak{u}(n)$.
Being $\{0\}$ and its complement in $\mathbb{C}^n$ the two orbits of $\mathbb{R}_+\times U(n)$, the equivariance of $\sigma$ implies that it induces a constant rank map from $\mathbb{C}^n\setminus\{0\}$ onto $\sigma(\mathbb{C}^n\setminus\{0\})$, which has to be an an orbit of $\mathbb{R}_+\times U(n)$ and so at least an initial submanifold of $\mathfrak{u}(n)$.
Being $\sigma^{-1}(\sigma(z))=\{e^{i\phi}z|\phi\in\mathbb{R}\}$ for any $z$, we get that $2n-1$ is the constant rank of $\sigma$ on $\mathbb{C}^n\setminus\{0\}$ and so even the dimension of $\sigma(\mathbb{C}^n\setminus\{0\})$ initial immersed submanifold of $\mathfrak{u}(n)$.
Finally by the observation of Jim Belk on the factorization of the action of $\mathbb{R}_+\times U(n)$ we have that $\sigma(\mathbb{C}^n\setminus\{0\}$ is an embedded submanifold of dimension $2n-1$.