Why should $\frac{2\pi}{5}$-rotations extend to symmetries of this platonically tesselated surface? I am reading in The geometry of Klein's Riemann surface, by Karcher & Weber. Section 3 starts off like this:

Let's try to construct a genus $2$ surface $M^2$ that is platonically tesselatied$^*$ by $F$ equilateral $\frac \pi 5 $-triangles. Such a triangulation has 3 vetrices, 15 edges and 10 faces. These ten triangles fit around one vertex to form a $\frac{2\pi}5$-decagon. What remains to be done is to give suitable identifications. 
We consider only identifications that satisfy necessary conditions for platonic tesselations. For example, we want the $\frac{2\pi}5$-rotations around the center of the decagon to extend to symmetries of the surface.

I don't see how the boldfaced part is deduced:


*

*Why would we only consider $\frac{2\pi}5$-rotations? $\frac \pi 5$-rotations should be symmetries just al well, right?

*What exactly does it mean that they extend to symmetries?


$^*$The authors say: "A tesselation of a Riemann surface is platonic if the symmetry group acts transitively on flags of faces, edges and vertices." They add: "Such a tesselation is also called a regular map."
 A: To answer your second question first, let $D$ denote the decagon, and $f : D \to M$ the quotient map obtained by the identifications. To say that a symmetry $\rho : D \to D$ extends to $M_2$ means that there exists a symmetry $\sigma : M \to M$ such that
$$f(\rho(x)) = \sigma(f(x)) \quad\text{for all $x \in D$}
$$
For the first question, it's possible that this is a typographical error, but to know whether it is or not would depend on information that your question does not provide (e.g. What is a platonic tesselation?). So I cannot answer the question. It is true that the $\pi/5$ rotation does not generate the full symmetry group of $D$ --- as you say, the $2 \pi / 5$ rotation is also a symmetry of $D$. Nonetheless the author's intent might be that the $2 \pi / 5$ rotation of $D$ and its powers will extend to $M$, but the $\pi/5$ rotation of $D$ will not extend to $M$.
I'll add, though, that if one uses the identification pattern that I know best, namely identifying opposite sides of the decagon $D$ to produce the genus 2 surface $M$, then the $\pi/5$ rotation of $D$ does indeed extend to $M$.
