The following Wikipedia's link https://en.wikipedia.org/wiki/Cantor_set affirms that "the Cantor set is a homogeneous space in the sense that for any two points $x$ and $y$ in the Cantor set $\mathcal {C}$, there exists a homeomorphism $ h:{\mathcal {C}} \mapsto {\mathcal {C}}$ with $h(x)=y$. These homeomorphisms can be expressed explicitly, as Möbius transformations."
Somebody knows this Möbius transformation?
I know a demonstration of the homogenity of Cantor set without using Möbius transformations.
That line in the Wikipedia page is nonsense. There are in fact only countably many Möbius transformations that map the Cantor set to itself (in fact, I suspect the only ones are $h(x)=x$ and $h(x)=1-x$). To prove this, suppose $h(x)=\frac{ax-b}{cx-d}$ is a Möbius transformation which maps the Cantor set to itself. Composing $h$ with $x\mapsto 1-x$ if necessary, we may assume $h$ is an orientation-preserving homeomorphism of $\mathbb{RP}^1$. But this means that $h$ preserves the cyclic ordering of any triple of points in $\mathcal{C}$. This means that if $x\in\mathcal{C}$ is the left endpoint of one of the triadic intervals used to construct $\mathcal{C}$, then $h(x)$ must also be such a left endpoint (since the left endpoints are exactly the points $x\in\mathcal{C}$ such that there exists $y\in\mathcal{C}$ such that there is no $z\in\mathcal{C}$ between $y$ and $x$). This means there are only countably many possible values of $h(x)$. Since a Möbius transformation is determined by its values on three points, we can determine $h$ by looking at its values on three left endpoints, and find there are only countably many possibilities for $h$.