 Solution

Use the inverse quantum circuit rule. Every gate in $C$ is self-inverse, so reverse their order. From left to right, the circuit for $V^{-1}$ is:

1. a controlled-NOT with the lower qubit controlling the upper;
2. $Z$ on the upper qubit and $H$ on the lower qubit;
3. a controlled-NOT with the upper qubit controlling the lower;
4. $H$ on the upper qubit.

This is the requested circuit description, with the two wires and control directions unchanged under each individual gate's adjoint.