First of all, if all three eigenvalues were really equal, then $\textbf{any}$ vector would be an eigenvector with that same eigenvalue. But that would also mean that the Hamiltonian is the identity matrix (up to an overall constant factor).
For the given Hamiltonian only one of the eigenvalues is $\lambda_1 = 2\cos\phi$. The other two are $\lambda_{2,3} = -\cos\phi \pm \sqrt{3}\sin\phi$. All of them can be determined from the usual equation defining the eigenvalues:
$$\det|H-\lambda I| = 0,$$
where $I$ is the $3\times 3$ identity matrix.
Knowing the eigenvalue, one can find the corresponding eigenvector by solving the linear system of 3 equations ($i$ could be 1, 2 or 3)
$$H v_i = \lambda_i v_i,$$
or more explicitly
$$\begin{pmatrix} 0 && e^{i\phi} && e^{-i\phi} \\ e^{-i\phi} && 0 && e^{i\phi} \\ e^{i\phi} && e^{-i\phi} && 0 \end{pmatrix} \begin{pmatrix} v_{i1} \\ v_{i2} \\ v_{i3} \end{pmatrix} = \lambda_i \begin{pmatrix} v_{i1} \\ v_{i2} \\ v_{i3} \end{pmatrix}.$$
To find the time-dependent solution one needs first to decompose the initial state into the eigenstates of the Hamiltonian, since each of them evolves in a known simple way. So we start start by finding a decomposition
$$v = \sum_{i=1}^3 c_i v_i$$
where $v$ is the initial state vector, $v_i$ are the Hamiltonian eigenvectors defined above, $c_i$ are complex numbers - coefficients of the decomposition. Eigenvectors of the Hamiltonian change with time by being multiplied by a time-dependent phase:
$$v_i(t) = e^{-\frac{i}{\hbar}\lambda_i t} v_i,$$
which follows from the time-dependent Schrodinger equation for the eigenvector:
$$i\hbar \frac{\partial}{\partial t} v_i(t) = \lambda_i v_i(t).$$
Finally, the time-evolved state $v(t)$ is given by the same linear combination of the time-evolved eigenstates as the initial state $v$ in terms of initial eigenstates $v_i$ which results in
$$v(t) = \sum_{i=1}^3 c_i e^{-\frac{i}{\hbar}\lambda_i t} v_i.$$
