The normal distribution pdf is defined to be:
$$ \phi(x| \mu, \sigma)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x-\mu)^{2}}{2\sigma^2}} $$
The cdf is given by:
$$ \Phi(x)=\int_{-\infty}^{x} \phi(t) dt $$
If we are given the value of $\phi(x)$ at three distinct points, this is sufficient to uniquely determine $\mu$ and $\sigma$.
The pdf of the general skew normal distribution is defined to be:
$$ f(x)=\frac{2}{\omega} \phi\left(\frac{x-\xi}{\omega}\right) \Phi\left(\alpha\left(\frac{x-\xi}{\omega}\right)\right) $$
How many points are needed to determine $\xi$, $\omega$ and $\alpha$?
Even though we have one more variable than in the non-skewed case, it looks like two skew normal distributions can only intersect in two places. So maybe we need only three again?
@Rahul points out in the comments that the pdf of two skew normal distributions can in fact intersect at three points. This leads to the guess, is four points sufficient to uniquely determine the distribution?