A rhombus $\mathcal R$ has short diagonal of length $1$ and long diagonal of length $2023$. Let $\mathcal R'$ be the rotation of $\mathcal R$ by $90^\circ$ about its center. If $\mathcal U$ is the set of all points contained in either $\mathcal R$ or $\mathcal R'$ (or both; this is known as the [i]union[/i] of $\mathcal R$ and $\mathcal R'$) and $\mathcal I$ is the set of all points contained in both $\mathcal R$ and $\mathcal R'$ (this is known as the [i]intersection[/i] of $\mathcal R$ and $\mathcal R'$, compute the ratio of the area of $\mathcal I$ to the area of $\mathcal U$.

[i]Proposed by Connor Gordon[/i]
1. **Substitution and Scaling**:
   - Let \( n = 2023 \) for simplicity.
   - Scale everything up by 2, noting that this will not affect the ratio of \(\mathcal{I}\) to \(\mathcal{U}\).

2. **Coordinate Setup**:
   - Define the vertices of \(\mathcal{R}\) as \((-1, 0)\), \((0, n)\), \((1, 0)\), and \((0, -n)\).
   - Define the vertices of \(\mathcal{R}'\) as \((0, 1)\), \((n, 0)\), \((0, -1)\), and \((0, -n)\).

3. **Intersection in the First Quadrant**:
   - By symmetry, the intersection point of the two sides of \(\mathcal{R}\) and \(\mathcal{R}'\) in the first quadrant will lie on the line \( y = x \).
   - The equation of one of the sides of \(\mathcal{R}\) is \( y = -\frac{1}{n}(x + 1) \).
   - Solving for the intersection point:
     \[
     y = x = -\frac{1}{n}(x + 1)
     \]
     \[
     nx = -x - 1
     \]
     \[
     (n + 1)x = -1
     \]
     \[
     x = -\frac{1}{n + 1}
     \]
     \[
     y = -\frac{1}{n + 1}
     \]
     Thus, the intersection point is \(\left(\frac{n}{n+1}, \frac{n}{n+1}\right)\).

4. **Area Calculation Using Shoelace Formula**:
   - The vertices of the intersection quadrilateral in the first quadrant are \((0, 0)\), \((0, 1)\), \(\left(\frac{n}{n+1}, \frac{n}{n+1}\right)\), and \((1, 0)\).
   - Using the Shoelace Formula to find the area:
     \[
     \text{Area} = \frac{1}{2} \left| 0 \cdot 1 + 0 \cdot \frac{n}{n+1} + \frac{n}{n+1} \cdot 0 + 1 \cdot 0 - (0 \cdot 0 + 1 \cdot \frac{n}{n+1} + \frac{n}{n+1} \cdot 1 + 0 \cdot 0) \right|
     \]
     \[
     = \frac{1}{2} \left| 0 + 0 + 0 + 0 - (0 + \frac{n}{n+1} + \frac{n}{n+1} + 0) \right|
     \]
     \[
     = \frac{1}{2} \left| -\frac{2n}{n+1} \right|
     \]
     \[
     = \frac{n}{n+1}
     \]
   - The total area of \(\mathcal{I}\) is \( 4 \cdot \frac{n}{n+1} \).

5. **Ratio Calculation**:
   - The area of \(\mathcal{U}\) is the area of \(\mathcal{R}\) plus the area of \(\mathcal{R}'\) minus the area of \(\mathcal{I}\):
     \[
     [\mathcal{U}] = 2 \cdot \left(\frac{1}{2} \cdot 2n \cdot 2\right) - 4 \cdot \frac{n}{n+1}
     \]
     \[
     = 2 \cdot 2n - 4 \cdot \frac{n}{n+1}
     \]
     \[
     = 4n - \frac{4n}{n+1}
     \]
   - The desired ratio is:
     \[
     \frac{[\mathcal{I}]}{[\mathcal{U}]} = \frac{4 \cdot \frac{n}{n+1}}{4n - \frac{4n}{n+1}}
     \]
     \[
     = \frac{\frac{4n}{n+1}}{4n - \frac{4n}{n+1}}
     \]
     \[
     = \frac{\frac{4n}{n+1}}{4n \left(1 - \frac{1}{n+1}\right)}
     \]
     \[
     = \frac{\frac{4n}{n+1}}{4n \cdot \frac{n}{n+1}}
     \]
     \[
     = \frac{4n}{4n \cdot \frac{n}{n+1}}
     \]
     \[
     = \frac{1}{n}
     \]

The final answer is \(\boxed{\frac{1}{2023}}\)