This is kind of a follow-up to a prior question on cyclic rotations of the qubits in an $n$-qubit register - the key to that question includes judiciously applying a number of SWAP gates. I'm interested in a small-sized reversible circuit to cyclically rotate the $n$ qubits, but to do so iff one of the qubits is $|0\rangle$ before the SWAPping.
I know I can copy the test qubit into an ancilla, and perform the rotation conditioned on this ancilla. But I have yet to find a way to assuredly uncompute this ancilla back to $|0\rangle$.
For $n=6$ here is an example Quirk circuit on a six-qubit register.
The bottom qubit is the ancilla, and the right-rotation is controlled on the third least significant qubit in the register being $|0\rangle$. In the left image, the rotation of the register occurs and the ancilla is properly uncomputed back to $|0\rangle$ when the third qubit is $|0\rangle$; in the right image, rotation does not occur when the third qubit is $|1\rangle$, but the ancilla stays as $|1\rangle$ when the third qubit is $|1\rangle$.
Is there a small reversible circuit that can unconditionally uncompute the ancilla (preferably growing polynomially with $n$)?
I'm motivated by considering a quantum walk on an old sliding puzzle I played with in the early 90's called "Spin-Out," which is isomorphic to the older "Chinese Rings" puzzle. In the image below, we can't slide to the left, because the spinner in the arrowed position is vertical, blocking the shift. And with some analogy to a CNOT gate, the spinner to the right of the arrowed spinner can spin because the arrowed spinner is vertical. A quantum walk with controlled-root-of-nots and controlled-root-of-shifts could be fun.
