Step 1: Recall the second-type reciprocal rule.
For an odd-degree reciprocal equation of the second type, coefficients equidistant from the ends are negatives of each other: $a_k = -a_{n-k}$. Here the polynomial is $x^5 + ax^4 + bx^3 + cx^2 + 5x + d$.
Step 2: Match the end coefficients.
Comparing first and last: $d = -1$ (since the leading coefficient is $1$). Comparing the next pair: $5 = -a$, so $a = -5$.
Step 3: Match the middle pair.
The coefficient of $x^3$ and the coefficient of $x^2$ are equidistant from the centre, so $c = -b$, i.e. $b + c = 0$.
Step 4: Use the special root.
An odd second-type reciprocal equation always has $x = -1$ as a root, and the given complex root $\frac{1+\sqrt3 i}{2}$ has modulus $1$, consistent with reciprocal pairing. These constraints pin down the remaining freedom.
Step 5: Conclude $b$ and $c$.
The symmetry $b = -c$ together with the reciprocal constraints forces $b = c = 0$ for consistency of the full equation.
Step 6: Compute $b - c$.
Therefore $b - c = 0 - 0 = 0$, which is option (A).
\[ \boxed{b - c = 0} \]