First of all, I'm going to assume your complete circuit looks like this:
[BTW, you should post your complete circuit if you expect to get any meaningful answers.]
Secondly, the unity voltage gain of the common collector refers to AC, not DC.
From the image above, you can see that the output voltage will be \$V_Z-V_{BE}\$.
And \$V_{BE}\$ will have some variation with the collector current, but not too much: \$V_{BE}\propto ln(I_C)\$.
On the other hand, \$I_B\$ is not negligible, it could be up to 20mA (for the transistor's minimum \$h_{FE}\$ of 50), and you don't really show how you are biasing your zener, so it could be that the base is sucking more than you are providing and the voltage across the zener will drop, and this drop will be directly reflected at the output voltage of this circuit.
By the way, from the 2SD1047 datasheet, \$V_{BE}\$ at 1A will be about 0.7V, so your output should be about 4.3V (not 5V), and like I said, will vary a bit with \$I_C\$. At 1A, it will dissipate quite a bit: \$1A(20V-4.3V)\approx 16W\$. The transistor should be able to thermally handle it though, since its thermal resistance is only 1.25°C/W.
In a common base circuit, the signal input source is applied to the emitter terminal. The "lump" of semiconductor that seperates base and emitter is a forward biased diode. The base is held at a constant voltage and therefore the emitter signal voltage causes a signal current to flow thru the base terminal.
Irrespective of whether the circuit is common emitter or base, the signal voltage across the forward biased B-E diode always produces a signal current through the base terminal. That signal current is determined by the diode biasing point i.e. how forward biased that junction is AND the magnitude of the signal.
That base signal current (even if the real input is the emitter) is amplified by the current gain and this results in a much larger signal current in the collector. The collector signal current (superimposed on the DC quiescent current of course) also flows thru the emitter terminal (plus of course the much smaller base signal current irrespective of where the signal source is connected).
So, an input connected to the emitter (common base) has to be able to cope with collector current AND the little bit of current that ends up flowing thru the base terminal.
Best Answer
You mean like this?
This is a push-pull emitter follower. The ideal case: the output equals the input. The practical case: there is a Vbe (~0.7V) crossover distortion which causes a deadband for input signals around 0V.