Absolute and relative tilings of the hyperbolic plane In Conway's Symmetries of Things on p. 265 I found these two tilings of the hyperbolic plane with the same vertex configuration $(3.5.3.5.3)$  (resp. vertex figure, as Conway calls it).

The difference lies in some permutation (of the vertices of one of the pentagons?) but I cannot see the effect of this permutation in the diagrams. I would have expected to see some crossing edges, but all I see are differently coloured pentagons – otherwise the two graphs look the same, i.e. isomorphic (what they presumably are not). What's going on here?

Daniel Huson sent me this plot in much better quality. It was made with Tegula. Thanks to Daniel.

 A: Let $T$ be the pictured tiling, where we forget about the colouring. Then its symmetry group is the maximal discrete subgroup of isometries of $\mathbb{H}^2$ which fixes $T$. Call that symmetry group $\Gamma$. This is the symmetry group of the picture on the right. It acts transitively on the vertices of $T$, and also acts transitively on the centres of the pentagons.
Now let $\Gamma^+$ be the index 2 subgroup of $\Gamma$ which contains the orientation-preserving isometries. Then $\Gamma^+$ also acts transitively on the vertices of $T$, but there are now two orbits for the centres of the pentagons. To see that, consider a pentagon and the 10 triangles nearest to it. On the orange pentagons, pairs of triangles point outwards in a anti-clockwise manner, while they point outwards in a clockwise manner on the yellow pentagons. Hence there is no orientation-preserving isometry taking an orange pentagon to a yellow one.
By absolute, Conway, Burgiel, and Goodman-Strauss mean the full symmetry group $\Gamma$ which acts transitively on the vertices of $T$. By relative, they refer to a proper subgroup of $\Gamma$ which still acts transitively on the vertices of $T$.
Let $e$ be an edge of $T$ which is shared by two adjacent triangles. Then there is a geodesic $m$ in $\mathbb{H}^2$ which contains $e$. A reflection in $m$ maps an orange pentagon onto a yellow one, and vice versa.
In the picture on the left, there are rotations of order 5 about the centres of the pentagons, and also rotations of order 2 about the intersection of any two mirror lines. These intersections are the midpoints of the edge $e$ and its translates by $\Gamma^+$. Hence the orbifold $\mathbb{H}^2/\Gamma^+$ is a $2$-sphere with cone points of orders 5, 5, and 2, and $\chi(\mathbb{H}^2/\Gamma^+)=\frac{-1}{10}$.
In the picture on the right, there are the same rotations, but now the rotation of order $2$ arises as the product of reflections in the two mirror lines passing through that point. So the orbifold $\mathbb{H}^2/\Gamma$ is a disk with one cone point of order 5 and a corner reflector of order 2. Thus $\chi(\mathbb{H}^2/\Gamma)=\frac{-1}{20}$. Then we can see that $\mathbb{H}^2/\Gamma^+$ double covers $\mathbb{H}^2/\Gamma$, which is to be expected since $\Gamma^+$ was an index 2 subgroup of $\Gamma$. This can also be seen by looking at the fundamental regions for the respective group actions.
