I know I can easily do this problem by drawing the possible necklaces, but I am trying to understand how to do it so I can do problems with way more possible arrangements. I am trying to use the following formula, in the context of group theory.
$$\# \mathrm{Orbits} = \frac{1}{|G|}\sum_{g\in G}|\mathrm{Fix}(g)|$$
Where $G$ acts on some set $X$, and $\mathrm{Fix}(g) = \{x \in X :g\cdot x=x\}$
For this case I know there are $\frac{4!}{2!2!}$ possible initial arrangements of the beads but after this I am lost.
If turning the necklace over is not allowed then the group of necklace transitions is rotating by 0,1,2 or 3 places, which is isomorphic to the cyclic group $C_4$.
There are 6 different sequences of two red beads and 2 blue beads:
rrbb, rbrb, rbbr, brrb, brbr, bbrr
All 6 sequences are fixed by rotating 0 places. None are fixed by rotating by 1 or 3 places (only a sequence with all beads the same colour could be fixed by these transitions). 2 sequences are fixed by rotating 2 places - rbrb and brbr.
So
$\# \mathrm{Orbits} = \frac{1}{|G|}\sum_{g\in G}|\mathrm{Fix}(g)| = \frac{1}{4}(6+2) = 2$
i.e. there are just two different necklaces with 2 red and 2 blue beads.
You can see this more directly by rotating a necklace until there is a red bead at the 12 o'clock position. Then the other red bead is either at 3 o'clock, 6 o'clock or 9 o'clock. The first and third cases are equivalent (up to rotation) to the sequence rrbb; the second case is equivalent (up to rotation) to the sequence rbrb.