The time-shifting property of the Fourier transform (see for example the Wikipedia Fourier Transform article) means that if $h(x) = f(x-x_0)$ then:
$$ \hat{h}(\xi) = \mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}\hat{f}(\xi) $$
If I expand out the right-hand side using the definition of $\hat{f}$:
$$ \mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}\hat{f}(\xi) = \mathrm{e}^{-\mathrm{i}2\pi x_0 \xi} \int_{-\infty}^{+\infty}f(x)\mathrm{e}^{-\mathrm{i}2\pi\xi x }\,\mathrm{d}x $$
I notice that the exponential term in front of the integral is not a function of $x$ so I bring it inside the integral:
$$ \begin{aligned} \mathrm{e}^{-\mathrm{i}2\pi x_0 \xi} \int_{-\infty}^{+\infty}f(x)\mathrm{e}^{-\mathrm{i}2\pi\xi x }\,\mathrm{d}x & = \int_{-\infty}^{+\infty}\mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}f(x)\mathrm{e}^{-\mathrm{i}2\pi\xi x }\,\mathrm{d}x \\ & = \int_{-\infty}^{+\infty}\left[\mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}f(x)\right]\mathrm{e}^{-\mathrm{i}2\pi\xi x }\,\mathrm{d}x \end{aligned} $$
Now I would like to claim that the final expression on the right-hand side is just $\hat{f}\left(\mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}f(x)\right)$ so that:
$$ \hat{h}(\xi)=\hat{f}\left(\mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}f(x)\right) $$
Then I would like to claim that if I take the inverse transform of both sides I get:
$$ h(x)=\mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}f(x)\quad (\, = f(x-x_0)) $$
But this seems to produce the wrong outcome when I calculate the modulus of $f(x-x_0)$. For a concrete example, let $f(x) = \mathrm{e}^{-x^2}$. Then:
$$|f(x)|^2 = (\mathrm{e}^{-x^2})^2$$
and
$$|f(x-x_0)|^2 = (\mathrm{e}^{-(x-x_0)^2})^2 = (\mathrm{e}^{{-x_0}^2+2xx_0})^2 (\mathrm{e}^{-x^2})^2 \ne |f(x)|^2 $$
but using the alternative expression I derived for $f(x-x_0)$:
$$|\mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}f(x)|^2 = |\mathrm{e}^{-\mathrm{i}2\pi x_0 \xi}|^2|f(x)|^2 = |f(x)|^2$$
What mistake am I making here?
I think the problem with interpreting the expression $\hat{f}(e^{-i2\pi x_0 \xi} f(x))$ is that $x$ is the variable in the time domain, not the frequency domain. There is no "$x$" in the expression for $\hat{h}(\xi)$. It disappears after the integration over $x$. (That's not the case with $x_0$ because it was a constant.)