I (think) I managed to prove that applying the Fourier integral transform to a function $f : \mathbb{R} \rightarrow \mathbb{C}$ twice results in $\mathcal{F}^2(f(x)) = f(-x)$. I then tried to prove the same thing for the DTFT, i.e. for $f : \mathbb{Z} \rightarrow \mathbb{C}$ but I got stuck. This is how far I got:
$$ \mathcal{F}^2(f(x)) = \mathcal{F}(\hat{f}(\omega)) = \sum_{\omega=-\infty}^{\infty}\hat{f}(\omega)e^{-i 2 \pi x \omega} $$
Since $\hat{f}(\omega)$'s domain is actually $[0, 1]$, should I be using an integral transform instead? The only thing I can think of other than that is that the $e$ term in the second transform is the complex conjugate of the $e$ term in the inverse transform.
https://en.wikipedia.org/wiki/Discrete-time_Fourier_transform
In the Wikipedia page, the DTFT of the sequence $(x[n])_{n=-\infty}^\infty$ is denoted by $X_{2\pi}(\omega)$, which stands for the function of a continuous variable $$\omega\in \mathbb R\mapsto X_{2\pi}(\omega)\in \mathbb C.$$ As such, if $\mathcal F$ denotes the mapping from $x$ to $X_{2\pi}$ (like in your question), then $\mathcal{F}^2$ makes no sense, as the range of $\mathcal F$ is different from its domain. This is something you already noticed in your question.
You can compose $\mathcal F_{DTFT}$ with $$ \mathcal F_{FS}(f)(n)=\int_0^1 f(\omega) e^{-i2\pi n\omega}\, d\omega$$ (FS stands for Fourier Series), which is a mapping of functions of a continuous variable into sequences. You should obtain $$\mathcal F_{FS}\mathcal F_{DTFT} x (n) = x(-n).$$