Is every arrangement reachable by shuffling this way? Suppose we have a vertical stack of $n$ distinguishable coins, each of which is either heads-up or tails-up. Let a shuffle be the following procedure: divide the stack at will into a top- and bottom-stack, and simply rotate the entire top stack $180^\circ$ as a unit. Thus, for $1\le k\le n$: 
$$\underbrace{x_1,...,x_{k}}_{\text{top stack}},\underbrace{x_{k+1},...,x_n}_{\text{bottom stack}} \\
\downarrow\\ \overline{x_k},...,\overline{x_{1}}, x_{k+1},...,{x_{n}}$$
where each $x_i$ is either $H_i$ or $T_i$, and
$$\overline{x_i} = 
\begin{cases} 
H_i,  & \text{if }x_i = T_i \\
T_i, & \text{if }x_i = H_i. 
\end{cases} $$
Since there are $2^n$ conceivable unsubscripted $H/T$ sequences, and there are $n!$ ways to append the subscripts to each, there are $n! \cdot 2^n$ conceivable ways to arrange the stack. 
Is it the case that, for any $n$, all of these conceivable arrangements are reachable by repeated shuffling (no matter what the initial arrangement)? 

Yes, you can get all of them. It suffices to show that you can get any desired coin to the bottom in either orientation, since you can then manipulate the rest of the stack without affecting the bottom coin and so build the desired sequence from the bottom up. But this is easy: any coin can be brought to the top in at most one move, where it can be flipped if necessary before the stack is inverted to bring it to the bottom.
