If: $y(t) = x(t)*h(t)$ and $g(t) = x(9t)*h(9t)$
(Where * is convolution)
How can I use properties of the Fourier transform to show:
$g(t) = Ay(Bt)$
and find constants?
I think A should be $1/81$ but I can only see that from first principles and not from a property.
For $a>0$,
$\mathcal{F}\{g(a t)\} = \int_{-\infty}^\infty g(a t) e^{- j \omega t} d t = \int_{-\infty}^\infty g(u) e^{-j \omega {u/a}} \frac{d u}{a} = \int_{-\infty}^\infty g(u) e^{-j (\omega/a) u} \frac{d u}{a} = G(\omega/a)/a$
via the $u$-substitution $u=at$, where $\mathcal{F}\{g(t)\} = G(\omega)$. The case for $a<0$ can be treated similarly.
Now use the convolution turning into multiplication by Fourier transform.