Let $d>1$ and $A$ be a diagonal matrix with real entries $\{ a_1, ...., a_d\}$ and $\det A \neq 0.$ For $m=(m_1,..., m_d) \in \mathbb R^d, Am= (m_1a_1,..., m_da_d)$ as usual. The Fourier transform of $f\in L^1(\mathbb R^d)$ is given by $\hat{f}(\xi) = \int_{\mathbb R^d} f(x)e^{-2\pi i x\cdot \xi} dx$
What relation one can expect between $\hat{f}(Am)$ and $\hat{f}(m)$?
Side note: In one dimension, if $h(x)=f(ax)$ $(x\in \mathbb R, 0\neq a \in \mathbb R),$ then $\hat{h}(\xi)= \frac{1}{|a|} \hat{f}(\frac{\xi}{a})$
If $A$ is an invertible matrix and $g(x)=f(Ax)$, then $\hat g(\xi)=\det(A)^{-1}\hat f(A^{-T}\xi)$. Here $A^{-T}$ is the transpose of the inverse (or the inverse of the transpose). To see this, simply write out the definition of $\hat g(\xi)$ and change the variable of integration from $x$ to $y=Ax$.
In your specific case the determinant is the product of all the diagonal entries and $A^{-T}=A^{-1}$.