Imagine you have a stack of $n$-even chips where the bottom half is blue and the top half is red. You split the stack equally and perform a perfect shuffle where the lowest blue chip remains on the bottom and hence the top red chip stays on the top. How many shuffles does it take to get all the blue chips on the bottom and all the red chips on top again?
One idea I had was to look at the order of the permutation at hand. Let $G$ be the permutation group of order $n$ with $\phi \in G$. Then, the permutation $\phi$ that mirrors the question is: $$\phi((1,2,3,\dots,(n-2),(n-1),n)) \to (1, (n/2+1),2,\dots,(n-1),(n/2),n)$$ I tried to find the order of $\phi$ by writing this in some general cyclic notation, but I couldn't seem to be able to figure it out. Also, the question has the subtlety that just the colors have to be reordered, not the actual original ordering of the chips. Any idea on how to go about solving this?
I don't think that one can restore the colours without actually restoring each chip to its original position.
Label the chips $0,1,\ldots,2n-1$. The shuffle fixes chips $0$ and $2n-1$ (forget these then) and moves any other chip $j$ to $2j$ considered modulo $2n-1$. So $k$ shuffles takes $j$ to $2^kj$ considered modulo $n$. So the smallest number of shuffles needed to restore all chips to the original position is the multiplicative order of $2$ modulo $2n-1$.
Suppose that one has just restored the colours. Write $A=\{1,2,\ldots,n-1\}$. Then $2^ka$ must lie in $A$ modulo $2n-1$ for all $a\in A$. In particular $2^k\equiv c\pmod{2n-1}$ where $c\in A$. We need to show that $c>1$ leads to a contradiction. Let $dc$ be the smallest multiple of $c$ that is $>(n-1)$. As $c>1$, $d\le n-1$ so $d\in A$, but as $(d-1)c\le n-1$ and $c\le n-1$, $dc\le 2n-2$. Thus this putative permutation maps $d\in A$ to $dc\notin A$ giving the required contradcition.