Find the total number of solutions to the following system of equations:

$ \{\begin{array}{l} a^2 + bc\equiv a \pmod{37} \\
b(a + d)\equiv b \pmod{37} \\
c(a + d)\equiv c \pmod{37} \\
bc + d^2\equiv d \pmod{37} \\
ad - bc\equiv 1 \pmod{37} \end{array}$
To find the total number of solutions to the given system of congruences, we will analyze each equation step-by-step.

Given system of equations:
\[
\begin{cases}
a^2 + bc \equiv a \pmod{37} \\
b(a + d) \equiv b \pmod{37} \\
c(a + d) \equiv c \pmod{37} \\
bc + d^2 \equiv d \pmod{37} \\
ad - bc \equiv 1 \pmod{37}
\end{cases}
\]

1. **Analyzing the second and third equations:**
   \[
   b(a + d) \equiv b \pmod{37}
   \]
   If \( b \not\equiv 0 \pmod{37} \), we can divide both sides by \( b \):
   \[
   a + d \equiv 1 \pmod{37}
   \]
   Similarly, for the third equation:
   \[
   c(a + d) \equiv c \pmod{37}
   \]
   If \( c \not\equiv 0 \pmod{37} \), we can divide both sides by \( c \):
   \[
   a + d \equiv 1 \pmod{37}
   \]

2. **Considering the case \( a + d \equiv 1 \pmod{37} \):**
   Substitute \( d \equiv 1 - a \pmod{37} \) into the first and fifth equations:
   \[
   a^2 + bc \equiv a \pmod{37}
   \]
   \[
   a(1 - a) - bc \equiv 1 \pmod{37}
   \]
   Simplify the fifth equation:
   \[
   a - a^2 - bc \equiv 1 \pmod{37}
   \]
   \[
   a - a^2 \equiv 1 + bc \pmod{37}
   \]
   Substitute \( a^2 + bc \equiv a \pmod{37} \) from the first equation:
   \[
   a - a^2 \equiv 1 + (a - a^2) \pmod{37}
   \]
   \[
   0 \equiv 1 \pmod{37}
   \]
   This is a contradiction, so \( a + d \not\equiv 1 \pmod{37} \).

3. **Considering the case \( b \equiv 0 \pmod{37} \) or \( c \equiv 0 \pmod{37} \):**
   If \( b \equiv 0 \pmod{37} \) and \( c \equiv 0 \pmod{37} \), the system simplifies to:
   \[
   a^2 \equiv a \pmod{37}
   \]
   \[
   d^2 \equiv d \pmod{37}
   \]
   \[
   ad \equiv 1 \pmod{37}
   \]
   The first two equations imply:
   \[
   a(a - 1) \equiv 0 \pmod{37}
   \]
   \[
   d(d - 1) \equiv 0 \pmod{37}
   \]
   Therefore, \( a \equiv 0 \) or \( 1 \pmod{37} \) and \( d \equiv 0 \) or \( 1 \pmod{37} \).

4. **Considering \( a \equiv 0 \pmod{37} \) or \( d \equiv 0 \pmod{37} \):**
   If \( a \equiv 0 \pmod{37} \), then \( ad \equiv 0 \not\equiv 1 \pmod{37} \), which is a contradiction.
   If \( d \equiv 0 \pmod{37} \), then \( ad \equiv 0 \not\equiv 1 \pmod{37} \), which is a contradiction.

5. **Considering \( a \equiv 1 \pmod{37} \) and \( d \equiv 1 \pmod{37} \):**
   If \( a \equiv 1 \pmod{37} \) and \( d \equiv 1 \pmod{37} \), then:
   \[
   ad \equiv 1 \cdot 1 \equiv 1 \pmod{37}
   \]
   This satisfies the fifth equation. Therefore, \( a \equiv 1 \pmod{37} \) and \( d \equiv 1 \pmod{37} \) are valid solutions.

Conclusion:
The only valid solution is \( a \equiv 1 \pmod{37} \), \( d \equiv 1 \pmod{37} \), \( b \equiv 0 \pmod{37} \), and \( c \equiv 0 \pmod{37} \).

The final answer is \(\boxed{1}\).