What is the correlation between the wave function of a photon and the vector potential? Sometimes (only when convenient) I hear professors and textbook writers considering the 4-vector potential $A_\mu=(\vec A,\phi)$ as the wave function of a photon. However, since photons have spin 1, I think the wave function of a photon should be a 3-vector.
For example, my professor of subnuclear physics used this argument to justify the fact that photons have parity $-1$: under inversion of parity the verse of currents changes and so does the magnetic field, which means $A_\mu$ becomes $-A_\mu$ under P transformations. 
 A: Contrary to a comment, there does exist a wave function for the photon because the quantized Maxell's equation, where the derivatives are interpreted as operators acting on the wave function represents the photon quantum mechanically.
In this link one can see the analogue between electron and photon wave equations;

In modern terms, a photon is an elementary excitation of the quantized electromagnetic field. If it is known a priori that only  one  such  excitation  exists,  it  can  be  treated  as  a  (quasi) particle, roughly analogous to an electron.

These are in E and B classical fields, but these  are connected with the four vector potential, as discussed here. 
It should be clear that an individual photon does not have an electric or magnetic field when measured. These reside in the complex valued wave function :

Note that it is a vector, because of the spin 1 as explained in the link.
It is the superposition of innumerable complex photon wave functions that build up real valued E and B fields of classical electromagnetism. Remember that to get a measurable observable one has to square this superposed huge wave function, so it is not a one to one correspondence from quantum to classical.
How the classical electromagnetic field emerges from a confluence of photons,  is shown, using the Quantum Filed Theory  formalism in this blog by Motl. 
The bottom line is that classical electromagnetism solutions work very well and in courses no emphasis is given to the underlying quantum mechanical wave function. of the photons.
A: $A_\mu$ is the solution of equation of motion from Maxwell's Lagrangian. These fields are some time also called "wave solutions" forexample see Mark Burgess Classical Covariant Fields equation 2.49, in momentum space we write
$$A_\mu(k)= C_k exp^{ik_\mu x^\mu}.$$
On the other hand the parity transformation you gave is wrong. This is a vector field which transform like a vector under parity transformation.
$$A_\mu=(\phi,\vec A)$$
after parity transformation
$$A^P_\mu= (\phi,-\vec A).$$
So $$A^P_\mu \not= -A_\mu$$
