In this paper we study the Hilbert transformations over ${{L}^{2}}(\mathbb{R})$ and ${{L}^{2}}(\mathbb{T})$ from the viewpoint of symmetry. For a linear operator over ${{L}^{2}}(\mathbb{R})$ commutative with the $ax\,+\,b$ group, we show that the operator is of the form $\lambda I+\eta H$, where $I$ and $H$ are the identity operator and Hilbert transformation, respectively, and $\lambda ,\eta $ are complex numbers. In the related literature this result was proved by first invoking the boundedness result of the operator using some machinery. In our setting the boundedness is a consequence of the boundedness of the Hilbert transformation. The methodology that we use is the Gelfand–Naimark representation of the $ax\,+\,b$ group. Furthermore, we prove a similar result on the unit circle. Although there does not exist a group like the $ax\,+\,b$ group on the unit circle, we construct a semigroup that plays the same symmetry role for the Hilbert transformations over the circle ${{L}^{2}}(\mathbb{T})$.