The correct answer is $\boxed{\frac{{\text{n}}}{2}}$ rpm.
Angular momentum is the rotational equivalent of linear momentum. It is a vector quantity, and its magnitude is given by the formula:
$$L = I\omega$$
where $I$ is the moment of inertia and $\omega$ is the angular velocity.
The moment of inertia of a circular disc is given by the formula:
$$I = \frac{1}{2}mr^2$$
where $m$ is the mass of the disc and $r$ is its radius.
The angular velocity of a circular disc is given by the formula:
$$\omega = \frac{2\pi}{T}$$
where $T$ is the period of rotation.
The angular momentum of a circular disc is therefore given by the formula:
$$L = \frac{1}{2}mr^2\frac{2\pi}{T} = \pi mr^2$$
The moment of inertia of a circular ring is given by the formula:
$$I = mr^2$$
The angular velocity of a circular ring is given by the formula:
$$\omega = \frac{2\pi}{T}$$
The angular momentum of a circular ring is therefore given by the formula:
$$L = mr^2\frac{2\pi}{T} = 2\pi mr^2$$
For the circular disc and the circular ring to have the same angular momentum, the mass and radius of the two objects must be the same. The angular velocity of the circular ring must therefore be half the angular velocity of the circular disc.
Therefore, the angular velocity of the circular ring is $\boxed{\frac{{\text{n}}}{2}}$ rpm.