$O$ is not dense in $\mathbb R^n$
Suppose $O=\{A^mx:m\in\mathbb N_0\}$ is dense in $\mathbb R$. There are two possibilities:
- $A^T$ has a real eigenvector $y$ corresponding to a real eigenvalue $\lambda$. Then \begin{aligned} &\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\ \Rightarrow&\ y^TO=\{\lambda^m y^Tx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}\\ \Rightarrow&\ Y=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}. \end{aligned}
- $A^T=A^\ast$ has an eigevector $z$ corresponding to some non-real eigenvalue $\overline{\lambda}$. Then the $\mathbb R$-linear span of the entries of $z$ must be two-dimensional and hence it is equal to $\mathbb C$. Therefore, \begin{aligned} &\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\ \Rightarrow&\ z^\ast O=\{\lambda^mz^\ast x:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}\\ \Rightarrow&\ Z=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}. \end{aligned}
It remains to show that $Y$ and $Z$ aren't really dense in $\mathbb R$ and $\mathbb C$ respectively. You may continue from here.