As Steve says, you need a boost converter.
Your comments about transistors amplifying current and FETs amplifying voltage are correct, but amplification relates to signals, which contain just information, and not a significant amount of _energy_ or power.
For example, you can use a transistor, FET or op-amp to amplify a tiny signal from a microphone or similar device, to a higher level to power a loudspeaker, but you'll need a power source for the amplifier. The actual energy that drives the loudspeaker comes from this power source; the input signal only determines how this energy is used. This is why a transistor radio needs batteries!
What you're talking about here is boosting the voltage of a power source, i.e. a source of energy. That is fundamentally different from amplification of a signal, which is what's being referred to in your comments about current and voltage gain in transistors, FETs and op-amps.
You need to understand first that power output, for any circuit, is never more than power input. In fact, power output is equal to power input multiplied by efficiency, which for a typical voltage converter is no more than 0.8 (80%).
So, if your solar cell can provide 1.5V at, say, 0.1A, its power output is 0.15 watts. This figure can be calculated from the power law: P = V I, where P is power in watts, V is voltage in volts, and I is current in amps. If you feed your solar cell's output into a boost converter with 80% efficiency, the available output power will be no more than 0.15 x 0.80, which is 0.12 watts.
If the boost converter has an output voltage of 5V, the maximum available current will be only 0.024A (24 mA). I calculated this from the P = V I formula rearranged to I = P / V where P is 0.12 watts and V is 5V. So in the process of boosting the voltage (which is what a "boost converter" does), you reduce the current. This is referred to in physics as "conservation of energy", aka "there's no such thing as a free lunch"
If your charger can only provide 24 mA, charging a cellphone battery of, say, 1200 mAh would take about 50 hours! Actually, it wouldn't work at all, because when you connect your cellphone, it will try to draw hundreds of milliamps, which will make the voltage converter's output voltage sag and disappear.
If you would like to imagine that you can get more power out of a circuit than you put into it, there are many YouTube channels featuring "over-unity" devices that claim to do this, and will gladly accept donations from gullible science-impaired folks ;-)
So your first priority is to get as much _power_ out of your solar cell as possible. If it's specified as 6V but only ever generates 1.5V, even in direct sunlight, I would return it to the supplier. Did you buy it through eBay? Oh well, you know not to do _that_ again! Many eBay sellers hawk cheap low-quality products with specifications that only loosely correlate with reality.
Check the specifications on your cellphone charger. Most likely it's something like 5V, 1A. This is 5 watts of power; somewhat higher than what your solar panel can provide, even in direct sunlight. The answer is to buy a larger solar panel from a reputable supplier that can actually provide 5W of power in reasonable lighting conditions.
