Shift-rotating random number generator
I think only for $n=1$.
The map $x\mapsto x\oplus\mathrm{rot}(x)\oplus s_0$ is 2-to-1 since $x$ and $x\oplus 1111\ldots 1$ both map to the same value. It follows that half of the values of $\{0,1\}^n$ do not lie in the image of this map. We can capture at most one missing value with our start point $0$, but the remaining $2^{n-1}-1$ points will never occur in the image.