It is well-known that the Fourier transform satisfies the following translation property: $$[\mathcal{F}(\mathcal{T}_hf)](\xi) = \text{e}^{2 \pi \text{i} \xi h} \cdot [\mathcal{F}(f)](\xi),$$ where $\mathcal{T}_h$ is the translation operator defined by $f(x) \mapsto f(x + h)$. By further defining the derivative operator as $$\mathcal{D} = \lim_{h \to 0} \frac{\mathcal{T}_h - \mathcal{I}}{h},$$ one can show that $$ [\mathcal{F}(\mathcal{Df})](\xi) = 2 \pi \text{i} \xi \cdot [\mathcal{F}(f)](\xi) $$ by linearity and continuity arguments. I'm wondering if it's also possible to prove the convolution theorem, just using the translation property of the Fourier transform? I know that in the finite dimensional case, a (circluar) convolution can be represented by a circulant matrix, which can in turn be written as a polynomial in simple translation matrices.
More generally, my question is:
Is the fundamental property of the Fourier operator that it diagonalizes translations?
That is, if we define $(\mathcal M_h \varphi)(\xi) = \text{e}^{2 \pi \text{i} \xi h} \cdot \varphi(\xi)$, is the Fourier operator $\mathcal{F}$ fundamentally characterized by $$ \mathcal{F} \circ \mathcal{T}_h = \mathcal M_h \circ \mathcal{F} \ ? $$