I posted this question on physics stack exchange but nobody answers it so hope to get an answer on math stack exchange,
What I understand after reading all answers from physics stack exchange related to residual gauge freedom and complete residual gauge fixing are as follows;
The gauge transformation is:
$A'_{\mu}$=$A_{\mu}$+$\partial_{\mu} f$, $f$ is any gauge function.
Here we can choose those four potentials only if it satisfies the Lorenz condition i.e $\partial^{\mu}$$A_{\mu}=0$, but the Lorenz condition is not a complete gauge condition since it doesn't give us a unique four-potential due to degrees of freedom in choosing gauge function $f$, as a result, there remains residual gauge freedom,
The degree of freedom of choosing the gauge function $f$ in the above gauge transformation gives us residual gauge freedom for $A_{\mu}$ so in order to choose a unique gauge $A_{\mu}$ after gauge transformation under Lorenz condition and simultaneously to reduce redundant degrees of freedom of $A_{\mu}$ due to gauge function $f$ we have to impose that the gauge function must satisfy homogeneous wave equation i.e Mathematically,
$\partial^{\mu}$$A'_{\mu}$=$\partial^{\mu}$$A_{\mu}$+$\partial^{\mu}$$\partial_{\mu}f$, Given that $\partial^{\mu}$$A'_{\mu}=0$ we get,
$0$=$\partial^{\mu}$$A_{\mu}$+$\partial^{\mu}$$\partial_{\mu}f$
-$\partial^{\mu}$$A_{\mu}$=$\partial^{\mu}$$\partial_{\mu}f$, again given that $\partial^{\mu}$$A_{\mu}=0$ we get,
$0$=$\partial^{\mu}$$\partial_{\mu}f$
$0$=$\square$$f$
$\square$$f$=$0$
i.e $\square$$f$=$0$ $\implies$ $\partial^{\mu}$$A'_{\mu}=0$.
So Lorenz condition $\partial^{\mu}$$A_{\mu}=0$ as well as $\square$$f$=$0$ completely fix the gauge by reducing the redundant degree of freedom. Is my understanding correct?
As you said, when imposing the Lorenz gauge, that is $\partial_\mu A^\mu = 0$, you still have some freedom in choosing a residual gauge $f$ which verifies $\Box f = 0$, to make sure that the gauge transformation $A_\mu \rightarrow A_\mu + \partial_\mu f $ does not spoil the Lorenz gauge.
What completely fixes the gauge in the end is choosing ONE particular $f$ that verifies $\Box f = 0$, and injecting the expression of $f$ in your gauge propagator and other non observable quantities which will depend on your choice of $f$.