Interesting and very valid question.
If you have 5V signal source with zero internal resistance loaded to antenna seen as 50 \$\Omega\$ resistor, then you pass power of 0.5Watt.
Say if you double the voltage using transformer 1:2, the power will be 2Watt. And so on.
In this case you do not need amplifier.
The practical problem is that the real signal source has non zero impedance. Typically 50 \$\Omega\$. So transformer 1:2 will make increase not in voltage terms but in impedance terms. The load will see the change as the source is now 100\$\Omega\$ of internal resistance.
So original voltage was 10V in the unloaded 50\$\Omega\$ source to reach 5V on 50\$\Omega\$ load.
Same 10V with transformed 1:2 load will see 50\$\Omega\$ load as if it was 25\$\Omega\$ load without transformer. The voltage will drop to 10*(25/75)=3.333. So power will be 0.44.. Watt instead of 0.5 watt. The loss of power as "reflected back" loss is caused by impedance mismatch.
Note that you can imagine the AC from wall outlet as as 2 stages DC power supply (that is a huge simplifcation but you may understand it better). Imagine that in a given moment the 2 holes in the wall give us a voltage difference of 120VDC between them, like a DC power supply of 120v (if exists). But in the next moment things get inverted. Now there is a voltage difference of -120VDC (minus) between the same 2 holes.
Now imagine that you took out your electronic device from the wall and inverted its plug in the wall (upside down). The only difference is that now in the first moment you will get -120v and in the second moment you will now get +120v. (changes in signal). But since this signal reversing is always happening many times per second, it doesn't matter if the first cycle is positive or negative. They will be switching forever.
As I said, this is a simple way to imagine things, because in the real world its a sinusoidal signal comming from the wall. But the main idea is that the voltage sometimes is negative and sometimes is positive in respect to earth. So you can imagine that the only difference that will exist when you change polarity of your electronic device plug will be a phase shift. But since we do not have an absolute notion of time, it doesn't matter (I mean, you can't make sure that always when you plug something in the wall, the first cycle will be positive or negative).
A rectifier may be built in many ways. But taking an example of a full bridge rectifier:
Note that in each cycle the current always goes from the top to the bottom of the resistor. So it does not matter if it's a positive or negative cycle comming from the wall, the main function of the rectifier bridge is precisely guarantee that the output signal preserve its polarity (DC) regardless of the AC polarity.
Just note that some electronic devices may damage if AC polarity is inverted. But this is a special case where you have a third wire (of ground) involved. Then you have a signal reference and maybe it's necessary that the AC polarity is correct so the device knows for sure which wire is neutral (ideally the same voltage of earth wire) and which wire is the phase.
Without that third wire, working with only neutral+phase (common AC outlet), your device cannot know which wire is which. It does only "see" that you are giving a voltage difference in its pins which vary from +120v to -120v. But it could be a pair of +240v and +120v (in respect to earth) or a pair of +120v and 0v (also in respect to earth). In both cases the difference is 120v and this is only thing your device "sees".
Best Answer
They have fixed voltage and current...
They don't! You have usually fixed voltage and a maximum limit for current. When a part of the phone needs more current, total consumption will increase. Since power is product of voltage and current, the power consumption will increase as well.
When you go over your maximum current limit, your voltage is going to drop. That's how it works in practice. If a system is well designed, then the maximum current limit will be high enough not to be reached in normal operation.