Let $f:\Bbb R^n\to \Bbb R^n$ be a continuous function such that $\int_{\Bbb R^n}|f(x)|dx\lt\infty$. Let $A$ be a real $n\times n$ invertible matrix and for $x,y\in\Bbb R^n$, let $\langle x,y\rangle$ denote the standard inner product in $\Bbb R^n$. Then
$$\int_{\Bbb R^n}f(Ax)e^{i\langle y,x\rangle}\,dx\overset{?}{=}\ \begin{array}{l} (1)\ \int_{\Bbb R^n} f(x) e^{i\langle(A^{-1})^T y,x\rangle} \frac{dx}{|\det A|} \\ (2)\ \int_{\Bbb R^n} f(x) e^{i\langle A^T y,x\rangle} \frac{dx}{|\det A|}\\ (3)\ \int_{\Bbb R^n}f(x)e^{i\langle(A^T)^{-1} y,x\rangle}\,dx \\ (4)\ \int_{\Bbb R^n}f(x)e^{i\langle A^{-1} y,x\rangle}\frac {dx}{|\det A|} \end{array}$$
Attempt:
Let $Ax=X\Rightarrow x=A^{-1}X$, then $dx=\frac {dX}{|\det A|}$. Using all these, we have:
$\int_{\Bbb R^n}f(Ax)e^{i\langle y,x\rangle}dx=\int_{\Bbb R^n}f(X)e^{i\langle y,A^{-1}X\rangle}\frac{dX}{|\det A|}=\int_{\Bbb R^n}f(X)e^{i\langle (A^{-1})^Ty,X\rangle}\frac{dX}{|\det A|}$.
Now if we again put $x=X\Rightarrow dx=dX$, then we get the last expression as,
$\int_{\Bbb R^n}f(x)e^{i\langle(A^{-1})^T y,x\rangle}\frac {dx}{|\det A|}$. That means 1) is true.