I am looking for a transformation $S$ of the Pauli matrices ($X,Y,Z$) such that
\begin{align*} S^{-1}XS=Y, S^{-1}YS=Z, S^{-1}ZS=X. \end{align*} Simply put, my question is a cyclic transformation of the matrices $X\to Y\to Z\to X$.
For example, Hadamard matrix $H=1/\sqrt{2}(X+Z)$ gives a transformation $X\leftrightarrow Z, Y\leftrightarrow -Y$ (I don't care the negative factor). If the transformation exists (or not), I want to know the proof or its mathematics behind it. Do you have any ideas?
I write here a detailed answer out of my comment. As is well known, the unitary one-parameter group $$e^{i \theta \vec{n}\cdot \vec{\sigma}/2}$$ has the property that $$e^{i \theta \vec{n}\cdot \vec{\sigma}/2} \sigma_k e^{-i \theta \vec{n}\cdot \vec{\sigma}/2} = \sum_{j=1}^3(R_{\vec{n}}(\theta))_{kj} \sigma_j\:, \tag{1}$$ where $R_{\vec{n}}(\theta)\in SO(3)$ is a spatial rotation around the unit vector $\vec{n}$ of an angle $\theta$.
As a consequence, let $\vec{n} = \frac{1}{\sqrt{3}}(\vec{e}_1 + \vec{e}_2+ \vec{e}_3)$ be the unit vector parallel to the diagonal of a cube with a vertex at $(0,0,0)$ and standard orthonormal axes $\vec{e}_j$. Then the action of a rotation $R_{\vec{n}}(2\pi/3)$ cyclically superposes the axes.
Taking (1) into account, we have that $$e^{i \theta \vec{n}\cdot \vec{\sigma}/2} \sigma_k e^{-i \theta \vec{n}\cdot \vec{\sigma}/2} = \sigma_{k'}\:, \quad e^{i \theta \vec{n}\cdot \vec{\sigma}/2} \sigma_{k'} e^{-i \theta \vec{n}\cdot \vec{\sigma}/2} = \sigma_{k''}$$ where $k,k', k''$ are a cyclic permutation of $1,2,3$.
$$e^{i \theta \vec{n}\cdot \vec{\sigma}/2}$$ can be explicitly computed by the formula (please check my numerical computation)
$$e^{i \theta \vec{n}\cdot \vec{\sigma}/2} = (\cos \theta) I + i (\sin \theta) (\vec{n}\cdot \vec{\sigma}) = -\frac{\sqrt{3}}{2} I + i \frac{1}{2\sqrt{3}} (\sigma_1 + \sigma_2+\sigma_3)\:.$$
