How can we show $\mathcal F^{-1}=\mathcal F^3$ using a specific definition of the Fourier transform $\mathcal F$? Let $d\in\mathbb N$ and $$\mathcal S(\mathbb R^d;\mathbb C):=\left\{f\in C^\infty(\mathbb R^d;\mathbb C):\partial^\beta f\text{ is rapidly decreasing for all }\beta\in\mathbb N_0^d\right\}.$$

I want to compute the inverse of the Fourier transform $\mathcal F$ restricted to $\mathcal S(\mathbb R^d;\mathbb C)$. My problem is that I'm only able to prove $$\mathcal F^{-1}=\mathcal F^3\tag1$$ if I'm using a specific convention for the definition of $\mathcal F$.

To be precise, I'm able to prove $(1)$ if I define $$(\mathcal Ff)(\omega):=\frac1{(2\pi)^{\frac d2}}\int f(x)e^{-{\rm i}\langle \omega,\:x\rangle}\:{\rm d}x\tag2\;\;\;\text{for }\omega\in\mathbb R^d,$$ since we then can show that $$(\mathcal F^2f)(\omega)=f(-\omega)\;\;\;\text{for all }\omega\in\mathbb R^d.\tag3$$
However, the definition I want to use is $$(\mathcal Ff)(\omega):=\int e^{-{\rm i}2\pi\langle\omega,\:x\rangle}\;\;\;\text{for }\omega\in\mathbb R^d.\tag4$$ But if I didn't make any mistake, I then obtain $$(\mathcal F^2f)(\omega)=\frac1{(2\pi)^d}f(-\omega)\;\;\;\text{for all }\omega\in\mathbb R^d\tag5,$$ which then gives $$\mathcal F^4f=\frac1{(2\pi)^{2d}}f\tag6.$$

So, did I made any mistake? From this wikipedia article, it seems like we should have $(1)$ with my definition $(4)$ as well.

 A: Yes, you made a mistake. I don’t know where, because you never mentioned how you got that extra factor of $\frac{1}{(2\pi)^d}$. If you define $\mathcal{F}$ using $(4)$, and if you define
\begin{align}
(\tilde{\mathcal{F}}f)(\omega):=(\mathcal{F}f)(-\omega)=\int_{\Bbb{R}^d}f(x)e^{2\pi i \langle x,\omega\rangle}\,dx,
\end{align}
then it is possible to show that $\mathcal{F}$ and $\tilde{\mathcal{F}}$ map the Schwartz space onto itself, and that they are inverses of each other there. This is part of the content of the Fourier-inversion theorem. From here, it follows (by composing with $\mathcal{F}$ and the reflection map $R= [f\mapsto [\omega\mapsto f(-\omega)]]$) that $\mathcal{F}^2=R$, i.e $(\mathcal{F}^2f)(\omega)=f(-\omega)$, and hence $\mathcal{F}^4=I$.
Folland’s real analysis text, and Stein and Shakarchi’s books also use this same convention for the Fourier-transform, so you may want to consult them.
A: Let $\mathcal F$ be the Fourier transform in your equation $(4)$ and $\mathcal F_0$ be the convention in your equation $(2)$. Then
$$
\mathcal F(f)(y) = (2\pi)^{d/2}\,\mathcal F_0(f)(2\pi y) = (2\pi)^{-d/2}\,\mathcal F_0\big(f(\tfrac{x}{2\pi})\big)(y)
$$
Therefore, using first the third expression and then the second
$$
\mathcal F^2(f)(x) = (2\pi)^{-d/2}\,\mathcal F_0\big(\mathcal F(f)(\tfrac{y}{2\pi})\big)(x) = \mathcal F_0\big(\mathcal F_0(f)\big)(x)
$$
hence
$$
\mathcal F^2 = \mathcal F_0^2.
$$
Therefore, if you know how to prove the Fourier inversion theorem for one of the versions, it works for the other as well ;)

(I suppose you forgot to divide $x$ by $2\pi$ in $(2\pi)^{-d/2}\,\mathcal F_0\big(f(\tfrac{x}{2\pi})\big)(y)$ in my first equation?)
