Four men are each given a unique number from $1$ to $4$, and four women are each given a unique number from $1$ to $4$. How many ways are there to arrange the men and women in a circle such that no two men are next to each other, no two women are next to each other, and no two people with the same number are next to each other? Note that two configurations are considered to be the same if one can be rotated to obtain the other one.
1. **Labeling and Initial Constraints**:
   - Denote the men as \( A, B, C, D \) and the women as \( a, b, c, d \).
   - Each man and woman pair share the same number: \( A \) and \( a \) have number 1, \( B \) and \( b \) have number 2, and so on.
   - The arrangement must alternate between men and women, and no two people with the same number can be next to each other.

2. **Arranging Men in a Circle**:
   - When arranging \( n \) distinct objects in a circle, the number of distinct arrangements is \((n-1)!\). This is because circular permutations consider rotations as identical.
   - For the 4 men, the number of ways to arrange them in a circle is \( (4-1)! = 3! = 6 \).

3. **Arranging Women Between Men**:
   - Given a fixed arrangement of men, there are 4 positions between them to place the women.
   - The women must also be arranged such that no two women are next to each other and no woman is next to a man with the same number.
   - For any fixed arrangement of men, there are 2 valid ways to arrange the women. For example, if the men are arranged as \( ABCD \), the women can be arranged as \( aBcAdCbD \) or \( aBdAcCbD \).

4. **Combining Arrangements**:
   - For each of the 6 ways to arrange the men, there are 2 ways to arrange the women.
   - Therefore, the total number of valid arrangements is \( 6 \times 2 = 12 \).

\[
\boxed{12}
\]