So, the problem is that you've got to enforce Fermionic antisymmetry, but Fock space tries to make things easier by making that invisible.
So if we've got two electrons in a box in a definite Fock state, the electrons definitively occupy some single-particle states which we can just call $1, 2$. The actual state that is being occupied is therefore:
$|\psi\rangle = |12\rangle - |21\rangle$
where the "first electron" (arbitrarily chosen) is in the first numbered state, etc.
Looking at your $C_a^\dagger$ and $C_a$ operators, it is somewhat clear that they are not capturing this distinction completely. Let us say that we're looking at $C_3^\dagger$ and $C_1$. Perhaps the action of $C_3^\dagger$ will look like:
$|123\rangle - |213\rangle - |132\rangle + |231\rangle - |321\rangle + |312\rangle$
Here I am associating the $+$ sign with appending onto the end, a $-$ sign with appending one before that. This means that $C_1$ should probably have a + sign for deleting from the end, a $-$ sign for deleting from the one before that, etc. This sign convention leads to the state:
$|23\rangle - |32\rangle $
But if we reverse these for $C^\dagger_3 C_1$ then the very same sign convention would force us first into the state $ -|2\rangle $ thus generating $ -|23\rangle + |32\rangle$. So you see that the results you get are negatives of each other, but this result is hidden by a naive Fock space solution.
We can focus on the orders which are associated with a + sign and phrase all of this simply as:
- For $C_1 C^\dagger_3$ I started with , prepended a 3 to get , swapped 1 to the front to get -, then removed the 1 from the front to get -.
- For $C^\dagger_3 C_1$ I started with , removed the 1 from the front, prepended with 3, got +.
Similarly with a starting point of three states, you start with  having a + sign associated with it:
- For $C_3 C^\dagger_4$ I started with , prepended a 4 to get , swapped 3 to the front with 3 swaps to get -, then removed it from the front to get -.
- For $C^\dagger_4 C_3$ I started with , swapped the 3 to the front with 2 swaps to get , removed the 3 from the front, prepended with 4, got +.
Now you can maybe see why they will always be negatives of each other: in the first case you will do $k$ swaps to get that number to the start of the permutation. In the second case you will do $k - 1$ because the 4 will not be there. So you'll do an odd number of swaps total.