Ok I did the problem again and I disagree. Here's my approach.
For the first transformation, any of the 11 options will achieve a satisfactory result (none of the 6 points has yet been transformed to). After that, the remaining 5 points can be transformed to in any order, as long as there is no repetition, so 5! = 120 orders for visiting points. At each step after the first, there are two transformations (one rotation and one reflection) that will get from one point to the next [distinct] point. Thus, the overall probability of success is
\(120\times\frac{11}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11} = \frac{42,240}{11^6}\)
so k = 42240
EDIT -------
I think you did 6! instead of 5!
Since you counted that the first transformation can be any of the 11 moves, then there are only 5! ways to arrange the other 5
