There is a bit of a contradiction hidden in the supposition of a transparent conducting material. Conducting materials, by their nature, are not transparent for the precise reason that they cannot support electric fields -- the electrons in the material will readjust to try and counteract any incoming electric field (light included). In the case of a perfect conductor the person inside would see nothing, all light would be reflected off the exterior surface.
Now let's dial back the "perfect conductor" condition a little bit. For a material with a finite conductivity electromagnetic radiation can penetrate into the material just a little bit (as the electrons can't adjust perfectly to compensate) and is exponentially attenuated the farther it goes into the material. The distance that it takes for the EM wave to be attenuated by a factor of $1/e \approx 0.36$ is called the skin depth of the material and is inversely related to the square root of the conductivity -- the less conductive the material the farther EM waves can propagate through it.
So, to answer your question: you would see an attenuated version of whatever is outside, the amount of attenuation depending on the thickness of the wall and the conductivity of the material.
I've (purposefully) left out a lot of details, and in general the effective conductivity depends on the wavelength of incoming radiation (it will be very different for microwave versus optical versus x-ray wavelengths) and it is a current active materials science problem to find (semiconductor) materials that have good conductivity at low frequency (so that they can be useful for electronics) but don't absorb light too much (at least when they are thin).