According to a time-frequency equivalency table in an undergraduate book on Harmonic Analysis, we have that:
or, in equation form:
$$ \widehat{\tau_h f}(\xi) = M_{-h} \widehat{f}(\xi). $$
Problem. This doesn't seem true to me.
Attempt. Recall that in the context of Fourier Analysis, the Fourier Transform $\widehat{f}(\xi)$ satisfies
$$ \widehat{f}(\xi) = \int_{- \infty}^{\infty} f(x) e^{-2 \pi i \xi x} dx $$
Then
$$ M_{-h} \widehat{f}(\xi) = e^{2 \pi i (-h) \xi} \left( \int_{- \infty}^{\infty} f(x) e^{-2 \pi i x \xi} dx \right) = \int_{- \infty}^\infty f(x) e ^{-2 \pi i \xi (x + h)} dx $$
and
$$ \widehat{\tau_h f}(\xi) = \widehat{f(\xi - h)} = \int_{- \infty}^{\infty} f(x) e^{- 2 \pi i x( \xi - h)} dx $$
so that
$$ \widehat{\tau_h f}(\xi) = \int_{- \infty}^{\infty} f(x) e^{- 2 \pi i x( \xi - h)} dx \ne \int_{- \infty}^\infty f(x) e ^{-2 \pi i \xi (x + h)} dx = M_{-h} \widehat{f}(\xi). $$
What am I missing?
EDIT. I'm now also having trouble understanding part (c) of this table. That is, I'm having trouble understanding why
$$ \widehat{M_h}f(\xi) = \tau_h \widehat{f}(\xi) $$
since
\begin{align*} \widehat{M_hf}(\xi) &= \widehat{e^{2 \pi i h \xi}f}(\xi) \\ &= \int_{-\infty}^{\infty} e^{- 2 \pi i \xi x} e^{2 \pi i h \xi} f(x) dx \\ &= \int_{-\infty}^{\infty} e^{- 2 \pi i \xi(x-h)} f(x) dx \\ \end{align*}
and
$$ \tau_h \widehat{f}(\xi) = \int_{- \infty}^{\infty} e^{- 2 \pi i (\xi - h)x} f(x) dx $$
so that it seems that
$$ \widehat{M_h}f(\xi) = \int_{-\infty}^{\infty} e^{- 2 \pi i \xi(x-h)} f(x) dx \ne \int_{- \infty}^{\infty} e^{- 2 \pi i (\xi - h)x} f(x) dx = \tau_h \widehat{f}(\xi)? $$
As far as I can tell, I'm not making any commutativity errors here as I was above.
I think what you're missing is that translation and Fourier transform don't commute, namely $$\widehat{(\tau_hf)}(\xi)\neq(\tau_h \hat f)(\xi).$$
The notation $\widehat{(\tau_hf)}$ means that you should first translate the function $f$ and then take the Fourier transform. What you've done is first taken the Fourier transform and then translated it. The calculation should have gone $$\widehat{(\tau_h f)}(\xi)=\int f(x-h)e^{-2\pi\mathrm{i}\xi x}\mathrm dx=\int f(x)e^{-2\pi\mathrm i\xi(x+h)}\mathrm dx,$$ where for the last equality I used the change of variables $x\mapsto x+h$.