Let $A = \left\{ 1,2,3,4,5,6,7 \right\}$ and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function.  Find the remainder when $N$ is divided by $1000$.
1. **Understanding the Problem:**
   We need to find the number of functions \( f: A \to A \) such that \( f(f(x)) \) is a constant function. This means that for all \( x \in A \), \( f(f(x)) = c \) for some constant \( c \in A \).

2. **Analyzing the Function:**
   Let's denote the constant value by \( c \). Without loss of generality, we can assume \( c = 1 \) (since we can multiply by 7 later to account for all possible constants). Thus, we need \( f(f(x)) = 1 \) for all \( x \in A \).

3. **Properties of \( f \):**
   - Since \( f(f(x)) = 1 \), it implies that \( f(x) \) must map to 1 for all \( x \in A \).
   - Therefore, \( f \) must map some elements of \( A \) directly to 1, and the rest of the elements must map to elements that eventually map to 1.

4. **Constructing the Function:**
   - Let \( k \) be the number of elements in \( A \) that map directly to 1. Thus, there are \( \binom{6}{k} \) ways to choose these \( k \) elements (excluding 1 itself).
   - The remaining \( 6 - k \) elements must map to one of the \( k \) elements that map to 1. Each of these \( 6 - k \) elements has \( k \) choices.

5. **Counting the Functions:**
   - For each \( k \), the number of ways to construct such a function is \( \binom{6}{k} \cdot k^{6-k} \).
   - Summing over all possible values of \( k \) from 0 to 6, we get the total number of functions for the case where \( f(f(x)) = 1 \).

6. **Calculating the Total Number:**
   \[
   N = 7 \sum_{k=0}^{6} \binom{6}{k} k^{6-k}
   \]
   The factor of 7 accounts for the 7 possible values of the constant \( c \).

7. **Evaluating the Sum:**
   \[
   \sum_{k=0}^{6} \binom{6}{k} k^{6-k}
   \]
   This sum can be evaluated using combinatorial techniques or generating functions, but for simplicity, we can compute it directly.

8. **Computing the Sum:**
   \[
   \sum_{k=0}^{6} \binom{6}{k} k^{6-k} = 1 + 6 \cdot 1 + 15 \cdot 1 + 20 \cdot 1 + 15 \cdot 1 + 6 \cdot 1 + 1 = 64
   \]

9. **Multiplying by 7:**
   \[
   N = 7 \cdot 64 = 448
   \]

10. **Finding the Remainder:**
    \[
    448 \mod 1000 = 448
    \]

The final answer is \(\boxed{448}\)