Problem: Let $ f:\mathbb R^{n} \hookrightarrow \mathbb R^{n} $ be a continuous function such that $ \int_{\mathbb R^{n}} {|f(x)|} dx < \infty $ . Let $ A $ be a real $ n\times n $ invertible matrix and for $ x,y \in \mathbb R^{n} $ , let $ \langle x,y\rangle $ denote the standard inner product in $ \mathbb R^{n} $ . Then $ \int_{\mathbb R^{n}} {f(Ax) \; e^{i \langle y,x\rangle }} dx = $
$$(1)~~~ \int_{\mathbb R^{n}} {f(x) \; e^{i \langle( A^{-1})^{T} y , x \rangle}} \; \frac{dx}{|\det A|} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)~~~ \int_{\mathbb R^{n}} {f(x) \; e^{i \langle A^{T} y , x \rangle}} \; \frac{dx}{|\det A|} $$ $$(3)~~~ \int_{\mathbb R^{n}} {f(x) \; e^{i \langle( A^{T})^{-1} y , x \rangle}} \; dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)~~~ \int_{\mathbb R^{n}} {f(x) \; e^{i \langle A^{-1} y , x \rangle}} \; \frac{dx}{|\det A|} $$
My thoughts: Since $ A $ be a real $ n\times n $ invertible matrix, then $~Ax=z\implies x=A^{-1}z~$
$$\therefore~~\langle x,y \rangle=\left\langle \left(A^{-1}\right)^Tz,y \right\rangle$$
Also $~dx=d\left(A^{-1}z\right)=dz/{|\det A|}~$
$$\implies \int_{\mathbb R^{n}} {f(Ax) \; e^{i \langle y , x \rangle}}dx= \int_{\mathbb R^{n}} {f(z) \; e^{i \langle( A^{-1})^{T} y , z \rangle}} \; \frac{dz}{|\det A|}= \int_{\mathbb R^{n}} {f(x) \; e^{i \langle( A^{-1})^{T} y , x \rangle}} \; \frac{dx}{|\det A|} $$Hence option $(1)$ is only true option.
Please verify my solution for the above problem and let me know if there be anything wrong in my solution and also provide if there be any other solution in other way of the same problem.