In quantum gauge theories it is usual to fix the gauge with the equation $\partial^\mu A_\mu = 0$ where $A_\mu$ is the gauge connection. From this gauge fixing condition the remaining gauge degree of freedom can be determined.
Unfortunaitly, there is appearing a determinant factor (Faddeev-Popov determinant) in the Feynman path integral coming from the change in the Haar measure due to this gauge fixing condition.
It is clear that without gauge fixing one integrates infinitely many (in the Feynman path integral) over the same contribution $e^{iS}$ since $S$ doesn't change under a gauge transformation. Is it possible to insert a gauge fixing factor $\delta(g-g_0)$ where $g$ is the gauge group and $g_0$ is a constant gauge group in the path integral, i.e. is it possible to restrict the gauge group to only a fixed function such that any other possibilities are not respected?
Or more precisely: What gauge fixing condition must be applied on $A_\mu$ to impose that the gauge group is fixed to only a single function over the spacetime?
EDIT (ATTEMPT): I am thinking about the gauge fixing condition $n^\mu A_\mu = 0$ with $n^\mu$ as a 4-vector function; can it satisfy above conditions? May be the gauge transformation given explicitely by $A_\mu \mapsto A_\mu + \Xi_\mu$ then it must also hold $n^\mu \Xi_\mu = 0$ and then if $\Xi^\mu$ is a fixed function, the 4-vector $n^\mu$ is chosen such that above gauge condition is satisfied.