1. Let the matrix be a $5 \times 5$ matrix with entries from $\{0,1\}$.

2. Given conditions:

   • Sum of elements in each row is $1$.
   • Sum of elements in each column is also $1$.

3. Since entries are only $0$ or $1$, the condition “sum of each row is 1” means:

   • In each row, there is exactly one entry equal to $1$ and the remaining $4$ entries are $0$.

4. Similarly, “sum of each column is 1” means:

   • In each column, there is exactly one entry equal to $1$ and the remaining $4$ entries are $0$.

5. Therefore, such a matrix is exactly a permutation matrix of order $5$.

   • A permutation matrix is obtained by placing exactly one $1$ in each row and each column.

6. Count such matrices:

   • In the first row, the $1$ can be placed in any of $5$ columns.
   • In the second row, it must be in one of the remaining $4$ columns.
   • In the third row, one of the remaining $3$ columns.
   • In the fourth row, one of the remaining $2$ columns.
   • In the fifth row, the last remaining column.

   Hence total number is $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5! = 120.$

7. Checking options:

   • A: $125$ ❌
   • B: $150$ ❌
   • C: $225$ ❌
   • D: $120$ ✅

Therefore, the required number of matrices is $\boxed{120}.$