To avoid possible double counting and similar problems it is useful to reduce all sequences equivalent by cyclic permutation to a "primitive form". From all identical (by rotation) sequences the smallest one (if considered as a binary number) is chosen as the primitive form. Another requirement imposed onto the primitive form is that it cannot be reduced to repeated sequence of a smaller length (for example $0101$ is not primitive because it can be reduced to two $01$).
As an illustration the primitive sequences of length 4 are:
$$
0001,0011,0111.
$$
Let $A_n$ be the number of primitive sequences of length $n$. Each of the sequences can be cyclically shifted by up to $(n-1)$ bits to obtain a new unique arrangement. By construction no primitive sequence can be obtained from itself or from another one by such a cyclic shift. Thus we have:
$$
2^n=\sum_{d|n}d A_d.
$$
By Möbius inversion one obtains:
$$
A_n=\frac{1}{n}\sum_{d|n}\mu\left(\frac{n}{d}\right)2^d.
$$
This is however not yet the number we are looking for. So far only the primitive sequences of length $n$ were counted. To obtain the number of all sequences of length $n$ the primitive sequences of smaller lengths dividing $n$ (periodically completed to the length $n$) shall be added:
$$
C_n=\sum_{d|n}A_d=\sum_{d|n}\frac{1}{d}\sum_{t|d}\mu\left(\frac{d}{t}\right)2^t=\frac1n\sum_{d|n}\varphi\left(\frac{n}{d}\right)2^d,
$$
where $\varphi(n)$ is the Euler's totient function.
Referred to your example:
$$
C_5=A_1+A_5=2+6=8.
$$
By inspection $A_1=2$ counts the sequences $00000$ and $11111$.
