I must say, I lost track of what's the issue discussed here, and to say it friendly, the Wikipedia in this case is pretty incomplete (Pauli comes to my mind, who was much less friendly in such cases).
First of all, although it's not explicit in the usual notation of the local Maxwell's equations, they are relativistic equations, and they can be written covariantly, i.e., you can use the Maxwell equations to calculate what happens in this famous case using any frame of reference (it's what Einstein discusses in the introduction in his famous "electrodynamics of moving bodies" paper of 1905 as the motivation for his reanalysis of the space-time description necessary to make these Maxwell equations obeying the special principle of relativity).
That said, I think it's not a very simple problem though, because you need the full Maxwell equations, and it's not easy to find a formulation, where you can really do the calculation analytically to the end, but you can do it at least formally, i.e., just keeping the fields, currents etc. in general terms. It's also good to start by thinking about what can be concretely measured.
I'd say, we take the conductor as a conducting ring and having put a volt meter in series (comoving with the ring). Supposed the motion is slow enough such that the voltmeter can follow the changes of the fields, what's measured is the electromotive force along the ring. I'm using Heaviside-Lorentz units (which are much more convenient for relativistic arguments than the SI).
(a) Moving-ring frame
In this case we have a given magnetostatic field ##\vec{B}(\vec{x})## obeying the corresponding static Maxwell equations, ##\vec{\nabla} \cdot \vec{B}=0##, ##\vec{\nabla} \times \vec{H}=\vec{j}=0##, ##\vec{H}=\mu \vec{B}##, but that's not relevant for our question here.
Now we have the moving ring, described by ##C:\vec{x}=\vec{r}(t,\lambda)=\vec{r}_0(\lambda)+\vec{v} t##, where ##\lambda## is the parameter for the curve. Also let ##\vec{v}(t,\lambda)=\partial_t \vec{r}(t,\lambda)=\vec{v}=\text{const}##. Further let ##A## be an arbitrary surface with ##C=\partial A##. Now we use Faraday's Law in integral form to get the electromotive force. We need the magnetic flux through the moving surface with the ring as its boundary:
$$\Phi(t)=\int_{A} \mathrm{d} \vec{x} \cdot \vec{B}.$$
The time derivative is not 0 although the magnetic field is static, because ##A## is moving, i.e., we get for the electromotive force
$$\mathcal{E}=-\frac{1}{c} \dot{\Phi}(t) = \frac{1}{c} \int_{\partial A} \mathrm{d} \vec{x} (\vec{E}+\vec{v} \times \vec{B}) =\frac{1}{c} \int_{C} \mathrm{d} \vec{x} \vec{v} \times \vec{B}.$$
That's the reading of the volt meter when calculated in this frame of reference.
(b) Moving-magnet frame
That's slightly more complicated ;-)). We have to use the Lorentz transformation of the fields first. In this frame the em. field becomes time dependent and has both electric and magnetic components. As is derived in any complete textbook of electromagnetics (most elegantly using the four-tensor formalism and writing the em. field components in terms of the antisymmetric Faraday four-tensor ##F_{\mu \nu}##),
$$\vec{E}_{\parallel}'(t',\vec{x}')=\vec{E}_{\parallel}(t,\vec{x})=0, \quad \vec{E}_{\perp}'(t',\vec{x}') = \gamma [\vec{E}_{\perp}(t,\vec{x}) +\vec{\beta} \times \vec{B}(t,\vec{x}),$$
$$\vec{B}_{\parallel}'(t',\vec{x}')=\vec{B}_{\parallel}(t,\vec{x})=\vec{B}_{\parallel}(\vec{x}), \quad \vec{B}_{\perp}'(t',\vec{x}')=\gamma[\vec{B}_{\perp}(t,\vec{x})-\vec{\beta} \times \vec{E}(t,\vec{x})]=\gamma \vec{B}_{\perp}(\vec{x}).$$
Now
$$\vec{x}=\gamma(\vec{x}'+\vec{v} t') \qquad (*),$$
and since the area and its boundary don't move, we simply have
$$\mathcal{E}'=\int_{\partial A} \mathrm{d} \vec{x}' \cdot \vec{E}' = \int_{\partial A} \mathrm{d} \vec{x}' \cdot \gamma [\vec{\beta} \times \vec{B}(\vec{x})] = \frac{1}{c} \int_{\partial A} \frac{\mathrm{d} \vec{x}}{\gamma} \cdot (\gamma \vec{v} \times \vec{B})=\mathcal{E},$$
as it should be. In the prelast step we have used (*) to transform the line element ##\mathrm{d} \vec{x}'=\mathrm{d} \vec{x}/\gamma##. Note that this is the correct transformation since the integral is to be taken at constant coordinate time, ##t'##!
