If I have a scalar field \begin{align} f: \mathbb{R}^3 &\rightarrow \mathbb{C}\\ (x, y, z) &\mapsto f(x, y, z) \end{align}
We can define an operator $P$ that takes a function like $f$ and gives a new function $Pf$. It is clear that for scalar fields like $f$ we have
$$ Pf(x, y, z) = f(-x, -y, -z) $$
But how is $P$ defined for vector fields? Suppose we have
\begin{align} \textbf{F}: \mathbb{R}^3 &\rightarrow \mathbb{R}^3\\ (x, y, z) &\mapsto F_x(x, y, z)\hat{\textbf{x}} + F_y(x, y, z)\hat{\textbf{y}} + F_z(x, y, z)\hat{\textbf{z}} \end{align}
Where each of $F_x, F_y$, and $F_z$ are scalar fields How is $P\mathbf{F}$ defined in this case?
$$ P\mathbf{F}(x, y, z) = \mathbf{F}(-x, -y, -z) = F_x(-x, -y, -z)\hat{\mathbb{x}} + F_y(-x, -y, -z)\hat{\mathbb{y}} + F_z(-x, -y, -z)\hat{\mathbb{z}} $$
Or do the basis vectors get transformed as well?
$$ P\mathbf{F}(x, y, z) = -\mathbf{F}(-x, -y, -z) = -F_x(-x, -y, -z)\hat{\mathbb{x}} - F_y(-x, -y, -z)\hat{\mathbb{y}} - F_z(-x, -y, -z)\hat{\mathbb{z}} $$
Or maybe just the vectors change and not the components:
$$ P\mathbf{F}(x, y, z) = -\mathbf{F}(x, y, z) = -F_x(x, y, z)\hat{\mathbb{x}} - F_y(x, y, z)\hat{\mathbb{y}} - F_z(x, y, z)\hat{\mathbb{z}} $$
I feel like the answer is going to depend on whether $\textbf{F}$ is a "axial" or "polar" vector like $\textbf{B}$ or $\textbf{E}$. But then I feel like the statement "$\textbf{B}$ is an axial vector" is more of a statement about how we choose to define some parity operator rather than something intrinsic about the magnetic field.