Under what conditions will a unitary matrix fix a subspace which does not diagonalize the generating Hamiltonian? Hello, this is my first post here.  I hope that it is not too vague; I will be as precise as I can, but I have more of a meta-problem so please forgive me if this is inappropriate.  My question is essentially of the form "what mathematical tools are most appropriate for investigating another problem?"  Of course, if it turns out that you can answer the underlying problem directly that would be great too :-)
Background and definitions
Given a Hermitian operator $H$, there exists a unitary operator $U(t)=\exp(-iHt)$ for any real $t$.
Let the set $\lbrace v_i\rbrace_{i=1}^N$ be a basis for the Hilbert space on which $H$ and $U$ act.
Let $V_n=\text{span}(\lbrace v_i\rbrace_{i=1}^n)$ be the span of a given subset of the full basis.
The question
Let $a\in V_n$.  In general, $U(t)\cdot a\notin V_n$, but it may be the case that for some $t_0$, $U(t_0)\cdot a\in V_n$.  I am looking for the more specific case in which this is true for all $a\in V_n$, for a single $t_0$.  So,

For a given $H$, what properties might determine whether or not there exists a $t_0$ such that $U(t_0)$ maps $V_n$ onto itself?  What tools would best tackle this problem?

It seems to me that this is equivalent to asking whether there is a $t_0$ for which $U$ is block-diagonalizable such that one of the blocks acts only on $V_n$.  I also think that my question may be related to orbits in a Lie algebra, but I am not certain how to rigorously state it in that language.
Additional information
Thanks to Helge, I have realized that I should have included a bit more motivation and am doing so here.
There are at least two ways in which the problem stated above could be answered trivially:


*

*Helge's solution of rational eigenvalues.

*$H$ could already be block-diagonal, so that the states of $V_n$ are never mapped out of it at any $t$.


I am interested in determining those $H$ that behave as specified for some $t_0$, but which do mix the states of $V_n$ with the rest of the Hilbert space for $t\in(0,t_0)$, and which result in $U(t_0)$ acting non-trivially on $V_n$.
 A: Let me just sketch some thing else. If $U(t_0) V = V$ for some subspace $V$ of the base Hilbert space, then we know that there exists an orthonormal basis of $V$ consisting of eigenvectors of $H$. This is the commutation result, I mentioned in the comments.
So, it clearly suffices to consider $H$ restricted to that subspace. The question that remains is how can it happen that $U(t_0)$ maps $n$ vectors into themselves, when $V$ is an $n$ dimensional space. If I am not totally mistaken this implies that $U(t_0)$ when restricted to $V$ is the identity.
Continuing my example. If the eigenvalues $E_1$ and $E_2$ satisfy $n E_1 = m E_2$ for some integers $n,m$, then we have that for any
$$
 v_1 = a_1 \psi_1 + b_1 \psi_2,\quad v_2 = a_2 \psi_1 + b_2 \psi_2
$$
the claimed property holds. Although not all eigenvalues are rational.

 Old Post
I am not sure I understand the question correctly, but here is one possible answer. Of course it is not the most general situation.
If all the eigenvalues of $H$ are rational then there exists a $t_0$ with the property you stated.
Proof
Let $E_1, \dots, E_N$ be the eigenvalues of $H$ and $\psi_1, \dots, \psi_N$ the corresponding eigenvectors. Then, we can write any $u$ as
$$
 u = \sum_{j=1}^{N} c_j \psi_j
$$
and
$$
 U(t) u =\sum_{j=1}^{N} e(- it E_j)    c_j \psi_j.
$$
Now assume that $E_j  = \frac{p_j}{q_j}$. Set
$$
 t_0 = 2 \pi \prod_{j =1}^{N} q_j
$$
Then we have $e(-i t_0 E_j) = 1$ for every $j$ and it follows that $U(t_0) u = u$ for any $u$.
A: OK, I think I have a comprehensive answer.  Since this is physics-inspired, I'll use bra-ket notation. 
First, in my comment above, you can see that if $U_{t_0}$ fixes a subspace $V_n \subset V$, then $U_{t_0}$ is block diagonal:

If we write $U$ in the basis $\{|v_1\rangle,...,|v_N\rangle\}$  and see how it acts on the basis of $V_n$, we must have the lower left block of $U$ be zero. Then, since the columns of unitary matrices form an orthonormal basis of the whole space, the columns of the upper left block of $U$ must be an orthonormal basis of $V_n$ . So, the columns of the right side of $U$ must span ${V_n}^{\perp}$, so that the upper right hand block is zero.

Next, let $W:=U_{t_0}|_{V_n}$ be the upper block of $U_{t_0}$ which acts on $V_n$:
$\quad U_{t_0} = \\left[ \begin{matrix} W & 0 \\\  0 & W' \end{matrix} \right]$
$W$ is unitary $n \times n$ matrix, and therefore can be diagonalized.  Without loss of generality, assume the orthonormal vectors $\{|v_1\rangle,...,|v_n\rangle\}$ diagonalize $W$ with respective eigenvalues $\{\exp(i w_i t_0)\}$.  Also, let $\{|h_1\rangle,...,|h_N\rangle\}$ be the orthonormal basis (in general, completely different than the $|v_i\rangle$) which diagonalizes $H$ with eigenvalues $h_j$.
Then, for all $i$ with $1 \le i \le n$,
$\quad \exp(i w_i t_0) = \langle v_i| U_{t_0}|v_i\rangle = \langle v_i|\exp(i H t_0) |v_i\rangle = \sum_{j=1}^N \exp (i h_j t_0) |\langle v_i | h_j \rangle|^2 $.
After multiplying both sides by $\exp(-i w_i t_0) $ we get
$\quad 1 = \sum_{j=1}^N \exp [i (h_j-w_i) t_0] |\langle v_i | h_j \rangle|^2$.
Now, since $\{|h_j\rangle\}$ is an orthonormal basis, the values $|\langle v_i | h_j \rangle|^2$ sum to unity (for fixed, normalized $|v_i\rangle$).  That means all the exponentials on the rhs for which $|\langle v_i | h_j \rangle|^2 \neq 0$ must be equal to unity. So,
$\quad (h_j-w_i) \frac{t_0}{2 \pi} \in \mathbb{Z}$
and
$\quad (h_j-h_{j'}) \frac{t_0}{2 \pi} \in \mathbb{Z} \quad \forall j,j'$ such that $\langle v_i | h_j \rangle $ and $ \langle v_i | h_{j'} \rangle \neq 0$.
In other words, for all $|h_j\rangle$ that have non-vanishing inner product with $|v_i\rangle$, the corresponding $h_j$ are separated by integer multiples of $\frac{2 \pi}{t_0}$.
It's easy to show that this is also a sufficient condition.  Given $H$ with eigenvalues $h_j$ and any subset $\mathbf{H}_{t_0} \subset \{h_j\}$ such that for all $h_j,h_{j'} \in \mathbf{H}_{t_0}$,
$\quad (h_j-h_{j'}) \frac{t_0}{2 \pi} \in \mathbb{Z}$
then any vector $|v\rangle$ in the span of eigenvectors corresponding to $\mathbf{H}_{t_0}$ will be an eigenvector of $U_{t_0} = \exp(i H t_0)$ with eigenvalue $\exp(i h_j t_0)$ (which is the same for all $h_j \in \mathbf{H}_{t_0}$).
(Everything below is old)
Example (Old)
Still not an answer, but here's a simple but non-trivial example: 
Let $U_t$ be the family of unitaries acting on $C^3$ which spatially rotate $R^3$ about the vector $(1,1,0)$ with period $T=2\pi$.  Let $t_0 = T/2=\pi$.  Then in the standard basis,
$\quad H = \frac{i}{\sqrt{2}}\left[ \begin{matrix} 0 & 0 & 1 \\\  0 & 0 & -1 \\\  -1 & 1 & 0 \end{matrix} \right] \quad ,$
$\quad U_{t_0/2} = \frac{1}{2}\left[ \begin{matrix} 1 & 1 & -\sqrt{2} \\\  1 & 1 & \sqrt{2} \\\  \sqrt{2} & -\sqrt{2} & 0 \end{matrix} \right]\quad , \quad U_{t_0} = \left[ \begin{matrix} 0 & 1 & 0 \\\  1 & 0 & 0 \\\  0 & 0 & 1 \end{matrix} \right] \quad , \quad U_{T} = I,$
so $U_{t_0}V_2 = V_2$, where $V_2 = \mathrm{span}\{(1,0,0),(0,1,0)\}$.  The eigenbasis is $\{(1,-1,-i\sqrt{2}),(1,-1,i\sqrt{2}),(1,1,0)\}$ with respective eigenvalues $\{1,-1,0\}$ for $H$. 
Note that $H$ and $U_{t_0 /2}$ are not block diagonal in the standard basis, but $U_{t_0}$ is.  In particular, no subset of the eigenvectors of $H$ form an orthonormal basis of $V_2$.
Commutation (Old)
I think the confusion over when you can infer that $H$ is block diagonal comes from the non-uniqueness of matrix roots.  Positive matrices have unique positive roots, of course, but unitary matricies do not have unique unitary roots.  For example, the $2 \times 2$ identity matrix has itself and
$\quad \left[ \begin{matrix} 0 & 1  \\\  1 & 0 \end{matrix} \right]$
as square roots, but only the latter mixes the first and second rows.  Likewise, I suspect that all non-trivial cases of unitaries fixing subspaces (where non-trivial is defined by Michael Underwood above) will have $t_0$ exactly such that non-degenerate eigenvectors of $H$ are degenerate for $U_{t_0}$.
