The question is in the title. To make it more precise, I am asking for the dimension of the set of (non-singular) normal endomorphisms of ${\bf K}^n$, seen as a submanifod $V_n$ of $GL(K^n)$, where $K={\bf R}$ or ${\bf C}$. (An endomorphism $A\in End(K^n)$ is normal if $AA^*=A^*A$, where $A^*$ denotes the adjoint of $A$ in the complex case, or its transpose in the real case.)
My guess is that the answer must be $n(n+1)/2$, but is there a simple proof, and could it be that the subspace in question is singular? Indeed, for $K={\bf C}$, $V_n$ is the image of the map $\phi:{\rm Diag}_n(K^*)\times O(K^n)$ defined by $\phi(D,U)=U^*DU$ (where I hope the notations are clear). This map is clearly differentiable but is also singular at $(D,U)=(I_n,I_n)$, where $I_n$ denotes the $n\times n$ identity matrix.