The question is inspired by physics applications, where we are often interested in diagonalizing a Hamiltonian (a Hermitian matrix) by a unitary transformation: $$ S^\dagger H S = \Lambda, $$ where $H$ is the Hamiltonian, $S$ is a unitary matrix and $\Lambda$ is a diagonal matrix. For silplicity I discuss further only the real Hamiltonians (i.e., symmetric matrices) and correspondingly orthogonal rotations, $S^{-1}=S^T$.
In 2-by-2 case one can solve directly for $S$ by representing it as $$ S=\begin{bmatrix}\cos\phi & \sin\phi\\-\sin\phi &\cos\phi\end{bmatrix} $$ and writing down the equations for the matrix elements.
For 3-by-3 and 4-by-4 Hamiltonians, the transformation $S$ can be formally found by solving the characteristic equation, $det(H-\lambda I)=0$ and calculating the eigenvectors. This transformations can be represented as a composition of Givens rotations, i.e., in 3-b-3 case: $$ S=S_z(\gamma)S_y(\beta)S_x(\alpha),\\ S_z(\gamma)=\begin{bmatrix}\cos\gamma & \sin\gamma & 0\\-\sin\gamma &\cos\gamma &0\\ 0 & 0 & 1\end{bmatrix}, S_y(\beta)=\begin{bmatrix}\cos\beta & 0 & \sin\beta\\0 & 1 & 0\\-\sin\beta & 0 &\cos\beta\end{bmatrix}, S_x(\alpha)=\begin{bmatrix}1 &0 & 0\\ 0 & \cos\alpha & \sin\alpha\\0 & -\sin\alpha &\cos\alpha \end{bmatrix}. $$
Can we solve for the angles $\alpha,\beta,\gamma$ analytically in 3-by-3 and 4-by-4 cases? On other words, is there such a representation of the solutions of cubic and quartic equations?