If I have $n$ concentric circles so that each section (annulus or a central disk) is divided into 4 pieces and allow the group $G= C_4 \times C_4 \times \ldots \times C_4$ to act on the set of colourings of the circles using $m$ colours by rotation of the annuluses.
Then how do I find the total number of fixed colourings by elements in $G$ so that I can use Burnside’s Lemma? It’s relatively easy to find the fixed number of colourings when an element of $G$ only rotates one annulus at a time however I’m not sure how to do this when more than one annulus isn’t acted on by an element.
I’ll assume that you meant “is acted on” in the last sentence.
The number of colourings fixed by an element $g$ of $G$ is the product of the numbers of colourings fixed by the components, since for each annulus you can independently choose a colouring that is fixed by the corresponding component of $g$; see multiplication principle.
Everything else in the problem also factors, so you’ll find that the number of equivalence classes of colourings under independent rotations of the annuli is the $n$-th power of the number of equivalence classes of colourings of each annulus under rotations. Here, too, you could have obtained this result directly by arguing that the equivalence classes of colourings under independent rotations of the annuli are formed by independently choosing an equivalence class for each annulus, and then applying the multiplication principle.