The older ones with substantial weight to them all contain transformers, so they will certainly be correctly isolated.
Most of the newer lightweight ones also contain small high-frequency transformers, so they also ought to be correctly isolated; however I would have my doubts about the very cheapest and possibly counterfeit "iphone charger" type of product.
If everybody is playing by the rules, anything with the "double-insulated" symbol (two nested square boxes) will be correctly isolated. Buy anything with this symbol from a reputable supplier and you should be good.
(For completeness : the mains isolation is usually "broken" by some small capacitors for noise suppression : these will (excepting counterfeits) be rated to high safety standards (Class Y or X2) and not considered to break the isolation for safety purposes; however there may be applications in e.g. sensitive medical electronics where even the tiny AC current they permit is too much. For all our sakes, I hope you're not asking about such applications on StackExchange! :-)
"Typical" fiber for communication is 9, 50, or 62.5 um diameter in the core, but there is a 125 um cladding that is also necessary for the fiber to work. There is also 900 um core plastic optical fiber. What is out there for illumination, I'm not sure. Clearly you're not working with one of the types that I'm familiar with.
But, the key point is that the core is smaller than the cladding. And the light coupled into the fiber will be (best-case) what falls on the core. With .87 NA you'll probably get pretty close to this ideal. So I'll just make up some numbers. Say you have a 30 um core in a 60 um cladding. Overall the maximum coupling efficiency you could expect from uniformly illuminating the end of the fiber bundle is 302/602 or 25%.
Then there's a factor for the packing density of round objects into a an area (the gaps between the fibers in the bundle) which I believe is about 78% best-case.
And a reflection loss of about 4% for light entering the glass from air.
Add all these up, (.25) x (.78) x (.96), and you have about 19% best-case coupling efficiency. (You'll need to re-calculate this knowing the correct factor for the first term)
If you are getting near this, I'd say you are actually doing pretty well.
Of course it wouldn't hurt to find an LED that emits in a narrowish cone instead of over half of all space, or even to add some kind of lens to focus the light on the area of the fiber bundle. But generally your best case coupling is still not going to be all that great.
Best Answer
I'm guessing that you'll be lucky to get 1% end-to-end efficiency with reasonably priced parts- maybe more like 0.1%. if that doesn't bother you maybe try 1W (or 3W) white LED or LEDs coupled to a fairly decent PV panel and some electronics to regulate the output to 3V.
There was a COTS fiber solution to this (a few hundred mW), but if I recall correctly price was several thousand dollars. IR lasers can have higher efficiency than LEDs, and they can be tuned to the PV wavelength.
Transformers are definitely going to be more efficient if you can find a way to use one. Unless you use light pipes or fibers to couple the light the large surface area of the emitter and PV cell may result in a significant coupling anyhow.