Symmetry in electricity and magnetism due to magnetic monopoles
Both of these follow from desirable properties of this hypothetical magnetic charge, namely:
- Magnetic charge is conserved.
- Magnetic field lines radiate outwards from positive magnetic charges.
- The net force between two magnetic charges moving at constant speed along parallel tracks is less than that between two stationary charges.
All three of these properties hold for electric charges. The last one may not be as familiar, but it basically works as follows: if we have a positive electric charge moving at constant velocity, it generates a magnetic field in addition to its electric field. A second positive electric charge moving parallel to the first one will therefore experience a magnetic force, and if you work out the directions, this force works out to be attractive. Thus, the net force between the two charges (electric and magnetic together) is less than the magnitude of the force they would exert on each other if they were at rest. [ASIDE: This can also be thought of in terms of the transformation properties of forces between different reference frames in special relativity, if you prefer to think of it that way.]
Now, the conservation of electric charge can be written in terms of the continuity equation: $$ \vec{\nabla} \cdot \vec{j}_e + \frac{ \partial \rho_e}{\partial t} = 0 $$ Note that this can be derived from Ampère's Law and Gauss's Law ($\epsilon_0 \vec{\nabla} \cdot \vec{E} = \rho_e$), using the fact that the divergence of a curl is always zero: $$ 0 = \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) = \mu_0 \vec{\nabla} \cdot \vec{j}_e + \mu_0 \frac{\partial ( \epsilon_0 \vec{\nabla} \cdot \vec{E})}{\partial t} = \mu_0 \left( \vec{\nabla} \cdot \vec{j}_e + \frac{\partial \rho_e}{\partial t} \right) $$ If we want to extend Maxwell's equations to magnetic charges, we need to have a magnetic version of Gauss's Law and add in a magnetic current term to Faraday's Law: $$ \vec{\nabla} \cdot \vec{B} = \alpha \rho_m \qquad \vec{\nabla} \times \vec{E} = \beta \vec{j}_m - \frac{\partial \vec{B}}{\partial t} , $$ where $\alpha$ and $\beta$ are arbitrary proportionality factors. But if we try to derive a continuity equation for magnetic charge from these two facts (as we did above for electric charge), we get $$ \beta \vec{\nabla} \cdot \vec{j}_m - \alpha \frac{\partial \rho_m}{\partial t} = 0, $$ and this is equivalent to the continuity equation if and only if $\alpha = - \beta$. Beyond this, the choice of $\alpha$ is to some degree arbitrary; different values correspond to different choices of which type of magnetic charge we call "positive", and what units we use to measure it. If we want to have magnetic field lines radiating away from "positive" magnetic charges, then we will want $\alpha > 0$; the usual choice in MKS units is to pick $\alpha = \mu_0$ (and $\beta = -\mu_0$), as you have in your equations above.
This negative sign in the magnetic current term Faraday's Law then implies that the electric field lines created by a moving magnetic charge will obey a "left-hand rule" instead of a "right-hand rule". In other words, the direction of $\vec{E}$ created by a moving magnetic charge would be opposite the direction of $\vec{B}$ created by a moving electric charge. If we still want two magnetic charges moving along parallel tracks to exhibit a lesser force than what they feel when at rest, then we must also flip the sign of the $\vec{v} \times \vec{E}$ term in the Lorentz force law to compensate for this flip.
You actually noticed something that is called electromagnetic duality. A duality correspond to two different theories which give the same physical results as long as we make specific mappings among their degrees of freedom. In the case of electromagnetism, what behaves as an electric field in one theory behaves as a combination of electric and magnetic fields in the dual one and vice-versa.
Indeed you noticed just a particular duality transformation, namely, $\vec E$ and $\vec B$ in the original theory behave as $-\vec B$ and $\vec E$, respectively, in the dual theory.
To see the general transformation we proceed as follows. The covariant formulation of Maxwell equations with sources, including magnetic monopoles, is $$\partial_\mu F^{\mu\nu}=j^\nu_e,\quad \partial_\mu \tilde F^{\mu\nu}=j^\nu_m,$$ where $F_{\mu\nu}$, $\tilde F_{\mu\nu}=\frac 12\epsilon_{\mu\nu\sigma\rho}F^{\sigma\rho}$, $j_e^\nu$ and $j_m^\nu$ are the electromagnetic tensor, the dual electromagnetic tensor, the electric four-current and the magnetic four-current, respectively. Those equations can be written in a single complex tensorial equation, $$\partial_\mu\mathcal F^{\mu\nu}=\mathcal J^\nu,\tag1$$ where $\mathcal F^{\mu\nu}=F^{\mu\nu}+i\tilde F^{\mu\nu}$ and $\mathcal J^\nu=j^\nu_e+ij^\nu_m$.
Eq. $(1)$ is invariant under the whole group of transformations $$\mathcal F^{\mu\nu}\rightarrow e^{i\varphi}\mathcal F^{\mu\nu},\quad \mathcal J^\mu\rightarrow e^{i\varphi}\mathcal J^\mu,\tag2$$ where $e^{i\varphi}$ is a complex phase, which implies $$E_i+iB_i\rightarrow e^{i\varphi}(E_i+iB_i),$$ or $$ \begin{align*} E_i&\rightarrow E_i\cos\varphi - B_i\sin\varphi ,\\ B_i&\rightarrow E_i\sin\varphi + B_i\cos\varphi , \end{align*},$$ and similar relations for the currents.
In particular, if we choose $\varphi=\pi/2$ then we get $$\vec E\rightarrow -\vec B,\quad \vec B\rightarrow \vec E.$$
Thus as long as the theory admits magnetic monopoles, there is a large freedom in what you call electric and magnetic fields.
Note that in vacuum, this electromagnetic duality holds despite the existence or not of magnetic monopoles.