Covariant Maxwell equations invariant under parity transformation I tried to proof that the Maxwell equations are invariant under parity transformations. Therefore I used the covariant formulation of the Maxwell equations 
\begin{align}
\partial_{\nu}F^{\nu\mu} &= \frac{4\pi}{c}j^{\mu}\\
\partial_{\nu}\tilde{F}^{\nu\mu} &= 0
\end{align}
and the parity transformation given by 
\begin{align}
P = \begin{pmatrix} 1 & 0 & 0 & 0  \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}
\end{align}
Regarding only the first equation $\partial_{\nu}F^{\nu\mu} = \frac{4\pi}{c}j^{\mu}$ we have
\begin{align}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{c}\frac{\partial}{\partial \text{t}} \\ \vec{\nabla} \end{pmatrix} = \begin{pmatrix} \frac{1}{c}\frac{\partial}{\partial \text{t}} \\ -\vec{\nabla} \end{pmatrix} 
\end{align}
as well as 
\begin{align}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} c\rho \\ \vec{j} \end{pmatrix} = \begin{pmatrix} c\rho \\ -\vec{j} \end{pmatrix} 
\end{align}
and 
\begin{align}
P \cdot F^{\nu\mu} &= \begin{pmatrix} 1 & 0 & 0 & 0  \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ E^1 & 0 & -B^3 & B^2 \\ E^2 & B^3 & 0 & -B^1 \\ E^3 & -B^2 & B^1 & 0  \end{pmatrix} = \begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ -E^1 & 0 & B^3 & -B^2 \\ -E^2 & -B^3 & 0 & B^1 \\ -E^3 & B^2 & -B^1 & 0  \end{pmatrix}
\end{align}
Based on these calculations, is there a way to see that Maxwell equations are invariant under parity transformations and if so how do I see it? 
 A: You are not applying the transformations correctly. Your transformation, $P$, is linear map that changes a vector into another vector. Well, $F^{\mu\nu}$ is a rank (2,0) tensor, not a vector (rank (1,0) tensor). This all becomes much clearer if you use index notation, rather than writing matricies. I will work in Cartesian basis.
So, let us denote your $P$ with a rank (1,1) tensor, $P^\mu_\nu$, such that for a vector $V^\mu=\left(V^0, V^1, V^2, V^3\right)^\mu$, the effect of $P$ would be:
$V^\mu\to\bar{V}^\mu = P^\mu_\nu V^\nu = (V^0, -V^1, -V^2, -V^3)$
Now the transformation for the electromagnetic tensor would be then:
$F^{\mu\nu}\to\bar{F}^{\mu\nu}=P^\mu_\eta P^\nu_\sigma F^{\eta\sigma}$
In this case, because your transformation is simple, you can rewrite this as a matrix equation (in general, it may not work so flawlesly):
$\bar{F}=P\cdot F \cdot P = \left(\begin{matrix}
    1       & 0 & 0 & 0  \\
    0       & -1 & 0 & 0  \\
    0       & 0 & -1 & 0  \\
    0       & 0 & 0 & -1  \\
\end{matrix}\right)
\left(\begin{matrix}
    0       & -E_x & -E_y & -E_z  \\
    E_x       & 0 & -B_z & B_y  \\
    E_y       & B_z & 0 & -B_x  \\
    E_z       & -B_y & B_x & 0  \\
\end{matrix}\right)
 \left(\begin{matrix}
    1       & 0 & 0 & 0  \\
    0       & -1 & 0 & 0  \\
    0       & 0 & -1 & 0  \\
    0       & 0 & 0 & -1  \\
\end{matrix}\right)=\left(\begin{matrix}
    0       & E_x & E_y & E_z  \\
    -E_x       & 0 & -B_z & B_y  \\
    -E_y       & B_z & 0 & -B_x  \\
    -E_z       & -B_y & B_x & 0  \\
\end{matrix}\right)$
So electric field transforms as a polar vector, whilst magnetic as axial. As is well known. You will also need to apply transform $J^\mu\to\bar{J}^\mu=P^\mu_\nu J^{\nu}$, and $\partial_\mu \to \bar{\partial}_\mu = \left(P^{-1}\right)^\nu_\mu \partial_\nu$ (i.e. you need to apply inverse transforms to co-variant tensors). When you do that, you will see that the equation for the new quantities will be the same as for the old ones, i.e. $\bar{\partial}_\mu \bar{F}^{\mu\nu}={4\pi}{c}\bar{J}^{\nu}$, so the equation does not change (vectors&tensors do)

Extra following the comment. 
I really advise against matrix notation in these calculations, so I will stop using it. Here is how you work out $\bar{F}^{\mu\nu}$:
$F^{0\{1,2,3\}}=\{-E_x, -E_y, -E_z\}$ 
$F^{12}=-B_z,\, F^{13}=B_y,\, F^{23}=-B_x$
So, using $P^0_0=1, P^1_1=P^2_2=P^3_3=-1$, and zero otherwise:
$\bar{F}^{0i}=P^0_0 P^i_j F^{0j} = (1)(-1)F^{0i}$, so electric field gets a minus
$\bar{F}^{ij}=P^i_s P^j_k F^{sk}=(-1)^2 F^{ij}$, so magnetic field is unchanged
Regarding the equation, the term on LHS goes as $\partial_{\mu}F^{\mu\nu}\to\left(P^{-1}\right)^{\kappa}_\mu\partial_{\kappa}P^\mu_\sigma P^\nu_\rho F^{\sigma\rho}=\delta^\kappa_\sigma\partial_{\kappa} P^\nu_\rho F^{\sigma\rho}=P^\nu_\rho \partial_{\mu}  F^{\mu\rho}$. You will find that the RHS transforms the same way, so the effect of transformation is to multiply the whole equation by an invertible operator $P$. Thus equation does not change.
A: As current is a vector, it is not invariant under parity. Therefore neither is Ampère's law. 
