I have the question:
The symmetry of a regular pentagon has 10 elements. Use 4 colours to colour the 5 regions of the pentagon below. Determine what the total number of distinct colourings of the pentagon using Burnside's lemma/counting theorem. (Two colourings are considered to be the same if there is a symmetry of the regular pentagon that maps one colouring to the other)
Is this how I would go about a question like this?
My solution.
I'll refer to each $|\operatorname{fix}(g)|$ by a particular letter. First, I have '$A$', any one of the 4 colours could be chosen for each of the 5 regions. This gives $4^5$.
Then I have the rotations around the centre point, '$B$'. There are 4 rotations ($\frac {1}{5}, \frac 25, \frac 35, \frac 45$), with each being able to one of the 4 colours. Therefore I have $4 \times 4$ possible choices.
Finally '$C$', I have the rotations in regards to 'cutting' through the edge. There are 5 edges. Each when 'cut' causes 2 'pairs' of opposite regions to swap/transpose. When transposed, these regions must be the colour which they were transposed with. So I have the axis (the cutting edge) which can be one of 4 colours and the each of the pairs which can also be one of the 4 colours. Therefore $C = 5 \times 4^3$
To find the number of colourings, the equation is: $$ \begin{align} \text{#Colourings} &= \tfrac {1}{10}[A + B + C] \\ &= \tfrac {1}{10}[4^5 + (4 \times 4) + (5\times 4^3)] \\ &= \tfrac {1}{10}[1360] \\ &= 136. \end{align} $$
Is this the correct method to go about a problem like this/have I missed anything?
Looks fine to me. That is indeed how you calculate the colorings modulo dihedral symmetry. One thing you might want to change is to say ahead of time that you're splitting into three different cases, and you're using $A,B,C$ to denote the number of fixed points collected in the final sum from the cases respectively. (Took me a minute to figure out what $A$ was supposed to be in your argument.)