The problem asks for the number of generator matrices that generate codes monomially equivalent to a specific ternary linear code $C$.

The logic is as follows:

1. Count the number of codes monomially equivalent to $C$: Let $\mathcal{C}_{equiv}$ be the set of codes monomially equivalent to $C$. This is the orbit of $C$ under the action of the monomial group $M_n(\mathbb{F}_q)$. The size of the orbit is given by the Orbit-Stabilizer Theorem: [ |\mathcal{C}_{equiv}| = \frac{|M_n(\mathbb{F}_q)|}{|Aut(C)|} ] Here, $n=7$ and $q=3$. The monomial group $M_7(\mathbb{F}_3)$ has size $n! (q-1)^n = 7! \cdot 2^7$. The problem states that the order of the monomial automorphism group of $C$ is $|Aut(C)| = 24$.

2. Count the number of generator matrices for each code: For a fixed linear code $C'$ of dimension $k$, the number of generator matrices is the number of ordered bases of $C'$. This is equal to the size of the general linear group $GL_k(\mathbb{F}_q)$. Here, $k=4$ and $q=3$. The size is $|GL_4(3)| = (3^4 - 1)(3^4 - 3)(3^4 - 3^2)(3^4 - 3^3)$.

3. Total number of generator matrices: Since every code in $\mathcal{C}_{equiv}$ has the same dimension, the total number of generator matrices is: [ N = |\mathcal{C}_{equiv}| \times |GL_4(3)| ]

I will perform the calculation using Python.

[PYTHON] import math

Constants

n = 7 k = 4 q = 3 aut_c_order = 24

Size of Monomial Group M_n(F_q) = n! * (q-1)^n

monomial_group_size = math.factorial(n) * (q - 1)**n

Number of equivalent codes = |M_n| / |Aut(C)|

num_equiv_codes = monomial_group_size // aut_c_order

Size of GL_k(F_q) = Product_{i=0 to k-1} (q^k - q^i)

gl_k_size = 1 for i in range(k): gl_k_size *= (qk - qi)

Total generator matrices

total_matrices = num_equiv_codes * gl_k_size

print(f"{monomial_group_size=}") print(f"{num_equiv_codes=}") print(f"{gl_k_size=}") print(f"{total_matrices=}") [/PYTHON] [THOUGHT] The code calculated the values:

• Monomial group size: $7! \times 2^7 = 5040 \times 128 = 645120$.
• Number of equivalent codes: $645120 / 24 = 26880$.
• Size of $GL_4(3)$: $(80)(78)(72)(54) = 24261120$.
• Total matrices: $26880 \times 24261120 = 652138905600$.

The logic holds. The number of generator matrices producing codes monomially equivalent to $C$ is the number of such codes times the number of bases per code.

Final Answer seems to be 652138905600. [/THOUGHT]

\boxed{652138905600}

The number of codes monomially equivalent to $C$ is given by the size of the orbit of $C$ under the monomial group action. The size of the monomial group $M_n(\mathbb{F}_q)$ for $n=7$ and $q=3$ is: [ |M_7(\mathbb{F}_3)| = 7! \times (3-1)^7 = 5040 \times 128 = 645120 ] Given that the order of the automorphism group of $C$ is 24, the number of distinct codes monomially equivalent to $C$ is: [ \frac{|M_7(\mathbb{F}_3)|}{|\text{Aut}(C)|} = \frac{645120}{24} = 26880 ] Each such code is a subspace of dimension $k=4$. The number of generator matrices for a specific code of dimension $k$ is the number of ordered bases for that code, which is equal to the order of the general linear group $GL_k(\mathbb{F}_q)$. For $k=4$ and $q=3$: [ |GL_4(3)| = (3^4 - 1)(3^4 - 3)(3^4 - 3^2)(3^4 - 3^3) = (80)(78)(72)(54) = 24261120 ] The total number of generator matrices producing codes monomially equivalent to $C$ is the product of the number of such codes and the number of generator matrices per code: [ 26880 \times 24261120 = 652138905600 ]

\boxed{652138905600}