Colin has a peculiar $12$-sided dice: it is made up of two regular hexagonal pyramids. Colin wants to paint each face one of three colors so that no two adjacent faces on the same pyramid have the same color. How many ways can he do this? Two paintings are considered identical if there is a way to rotate or flip the dice to go from one to the other. Faces are adjacent if they share an edge.
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1. **Understanding the Problem:**
   Colin has a 12-sided die made up of two regular hexagonal pyramids. He wants to paint each face one of three colors such that no two adjacent faces on the same pyramid have the same color. We need to find the number of ways to do this, considering that two paintings are identical if they can be rotated or flipped to match each other.

2. **Symmetry Considerations:**
   The die has 12 symmetries: 6 rotations and 2 flips. We will use Burnside's Lemma to count the number of distinct colorings.

3. **Counting Colorings for the Identity Symmetry:**
   Let \( F_n \) be the number of ways to color a single \( n \)-gonal pyramid. We need to find \( F_6 \) for a hexagonal pyramid. The recurrence relation for \( F_n \) is given by:
   \[
   F_n = F_{n-1} + 2F_{n-2}
   \]
   with initial conditions \( F_2 = 3 \) and \( F_3 = 3 \).

4. **Solving the Recurrence Relation:**
   We solve the recurrence relation:
   \[
   F_n = 2^n + 2 \cdot (-1)^n
   \]
   For \( n = 6 \):
   \[
   F_6 = 2^6 + 2 \cdot (-1)^6 = 64 + 2 \cdot 1 = 66
   \]
   Therefore, the number of ways to color a single hexagonal pyramid is 66. Since the die consists of two such pyramids, the total number of ways to color the die without considering symmetries is:
   \[
   66^2 = 4356
   \]

5. **Applying Burnside's Lemma:**
   We need to consider the effect of each symmetry on the colorings:
   - **Identity Symmetry:** All 4356 colorings are fixed.
   - **Rotations by \( \pm 60^\circ \):** These rotations do not preserve the coloring because each face must map to a face of the same color, which is not possible with the given constraints.
   - **Rotations by \( \pm 120^\circ \) and \( \pm 180^\circ \):** Each of these rotations fixes 36 colorings.
   - **Flips:** Each flip fixes 66 colorings because the top and bottom pyramids are identical up to rotation.

6. **Calculating the Average Number of Fixed Colorings:**
   Using Burnside's Lemma, the average number of fixed colorings is:
   \[
   \frac{1}{12} \left( 4356 + 3 \cdot 36 + 6 \cdot 66 \right)
   \]
   Simplifying the expression:
   \[
   \frac{1}{12} \left( 4356 + 108 + 396 \right) = \frac{1}{12} \left( 4860 \right) = 405
   \]

The final answer is \( \boxed{405} \).