According to my solution there should be more fixed points...

Question

I have solved the following exercise:

A group of order $55$ acts on a set of order $18$. Then there are at least $2$ fixed points.

But according to my solution, there should be at least 3 fixed points (I solved it via a form of Burnsides lemma and some basic number-theoretic reasoning). Is that true ?

Answer 1

There might be an orbit of length 11, and one of length 5, so two fixed points are definitely a possibility.

For an example, take the cyclic group generated by the product of two disjoint cycles of length 5 and 11 in the symmetric group $S_{18}$.