This is an answer check for the number of distinct colouring's for the regular pentagon given only four colour choices.
I have the rotational group action $\{\frac15,\frac25,\frac35,\frac45,\frac55=0\}$ and the symmetric flip from each vertex through the middle of the opposing fifth section.
If this second group action isn't clear, I will upload a picture perhaps. We are looking at the colours of each of the five edges(or if you would rather visualise five segments cutting the pentagon's area, that would be equivalent). Where two permutations are equivalent.
I have completed this problem using the counting theorem, e.g:
$t = \frac{1}{|G|}\sum \limits_{g\in G} |\text{fix}(g)|$
This yielded the answer $136$
Note this is an assigment question, hence a yes/no answer is preferred!
As you say, your group $G$ is generated through the 5 rotations and a reflection, yielding $\left| G\right| = 10$, which can be thought of as the group of 5 rotations and 5 reflections through the various vertex-to-opposing-edge axes.
The total number of edge colourations is $4^5$.
The identity element of $G$ fixes all $4^5$ colourations.
The $4$ partial rotations fix only $4$ colourations (namely, those where all five edges have the same colour).
The 5 reflections each fix $4^3$ colourations (namely, those where the two edges meeting at the axis vertex have equal colour and the two edges adjacent to the axis edge have equal colour).
Thus, by the counting lemma (which you're correct to employ): $$n_{\rm orbit} = \frac{1\cdot 4^5 + 4\cdot 4 + 5\cdot 4^3}{10} = \frac{1360}{10} = 136$$ Hence yes, your answer is correct.