Will a vector rotate back to itself under unitary rotation?

Question

Let there be an arbitrary complex vector $\vec{v} \in \mathbb{C}^D$ that has unit length defined as $\sum_i v_i v_i^* = 1$. Let there be a Hermitian $D \times D$ matrix $H$. Given arbitrary $\vec{v}$ and $H$, will there always exist a non-zero $t \in \mathbb{R}$ such that $e^{iHt} \vec{v} = \vec{v}$?

Answer 1

No. Consider $H=\begin{pmatrix} 1&0\\0&\sqrt3\end{pmatrix}$. Then $$e^{iHt}=\begin{pmatrix}e^{it}&0\\ 0& e^{i\sqrt3t}\end{pmatrix}$$

And it is now apparent that the vectors $v\in\Bbb C^2$ such that there is some $t\ne0$ satisfying $e^{iHt}v=v$ are the ones such that either $v_1=0$ or $v_2=0$.

Added: In the same spirit, it isn't hard to show that the hermitian matrices $H$ such that for all unitary vectors $v$ there is some $t\in\Bbb R\setminus \{0\}$ such that $e^{itH}v=v$ are exactly the ones such that there is some real number $\alpha$ such that $\frac\lambda\alpha\in\Bbb Q$ for all $\lambda$ eigenvalues of $H$.

More precisely, let $H=H^*\in\Bbb C^{n\times n}$ and $v\in \Bbb C^n$. Let's call $V_\lambda=\ker(H-\lambda I)$ and let's impose the following equivalence relation on $\operatorname{spec}H$: $$\lambda R\mu\iff \lambda=\mu=0\lor \left(\lambda\ne0\land\mu\ne0\land \frac\mu\lambda\in\Bbb Q\right)$$ Let's call $W_{\lambda}=\bigoplus\limits_{\mu R\lambda} V_\mu$. Then there is some $t\in\Bbb R\setminus\{0\}$ such that $e^{itH}v=v$ if and only if $v\in V_0+ W_\lambda$ for some $\lambda\in\operatorname{spec}H$.