Let $V=\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & 0 & 1 \\ 1 & 1 & -1 \\ -i & i & 0 \end{bmatrix}$. Some tedious work shows that $V^{-1}=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 & i \\ 1 & 1 & -i \\ 2 & 0 & 0 \end{bmatrix}$.
Then $\Lambda = V^{-1} A V = \begin{bmatrix} i & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & -1 \end{bmatrix}$. Solving $\dot{y} = \Lambda y$ gives $y(t) = e^{\Lambda (t-t_0)} y(t_0) =\begin{bmatrix} e^{i(t-t_0)} & 0 & 0 \\ 0 & e^{-i(t-t_0)} & 0 \\ 0 & 0 & e^{-(t-t_0)} \end{bmatrix} y(t_0)$.
Since $\dot{y} = \Lambda y$, we have $\dot{Vy} = V\Lambda y$, and letting $x = Vy$, we have $\dot{(V V^{-1} x)} = \dot{x} = V\Lambda V^{-1} x = A x$ (and $x(t_0) = V y(t_0)$, of course). Hence the solution is
$x(t) = V e^{\Lambda (t-t_0)} V^{-1} x(t_0)$
Note that $e^{\Lambda (t-t_0)} = e^{i(t-t_0)} e_1 e_1^T + e^{-i(t-t_0)} e_2 e_2^T +e^{-(t-t_0)}e_3 e_3^T$, we can compute $M_i = V e_1 e_1^T V^{-1}$, $M_{-i} = V e_2 e_2^T V^{-1}$ and $M_1 = V e_3 e_3^T V^{-1}$ to get
$x(t) = (M_i e^{i(t-t_0)} + M_{-i} e^{-i(t-t_0)} + M_1 e^{-(t-t_0)}) x(t_0)$
Grinding through the details gives $M_i=\frac{1}{2}\begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 & i \\ -i & -i & 1 \end{bmatrix},\ \ \ M_{-i} = \overline{M_i},\ \ \ M_1=\begin{bmatrix} 1 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Combining all of this together gives
$x(t) = \begin{bmatrix} e^{-(t-t_0)} & 0 & 0 \\ \cos(t-t_0)-e^{-(t-t_0)} & \cos(t-t_0) & -\sin(t-t_0) \\ \sin(t-t_0) & \sin(t-t_0) & \cos(t-t_0) \end{bmatrix} x(t_0)$