What does this transistor circuit do?
This circuit is a capacitance multiplier designed to remove ripple from a 'dirty' regulated power supply. The value of the capacitance seen from the output terminals is a multiple (not exactly beta times) of the capacitance seen from emitter and base.
In its more 'general form' it allows for a re-scaling of the output voltage via a voltage divider, like this:

Here I added a 0.2 Vpp 50 Hz ripple to the 9V input voltage and I get 6.3 V output with just 0.5 mVpp ripple superposed (this is an old simulation I did years ago). Here is the result of the simulation from start up:

As you can see the output voltage is basically a flat line.
In your circuit the diode add a polarity inversion protection, so you will see a lower voltage, according to the current supplied.
The single resistor version
(edited after reading other answers and comments) To address the difference between the multiplier with R1 and R2 and your design, note that the capacitor sees a resistance across its terminal in your design as well. It's the resistance seen from the base of the transistor. That is enough to keep the transistor out of saturation (as long as the ripple is small, as it usually is).
Here is a simulation with a 6.3V input with ripple of 0.2 Vpp @ 50 Hz and a load of 220 ohm; I also added a Schottky diode as polarity inversion protection and to bring the voltage at almost exactly 5V.

Note that this design is highly dependent on the value of RL since the setpoint depends on the implicit voltage divider former by R1 and (basically) beta times RL.
As noticed in another answer, this circuit requires a bit of time to reach its steady state value, so that along with ripple suppression it adds a soft-start functionality. This might be an intended function (or even the intended function) of your circuit. The simulation from t=0 shows that:

But even with a single resistance, a capacitance multiplier still works well as a ripple suppressor circuit if the load is low enough (more on this later) and the ripple is reasonably small. Here is the output of the same simulation a few seconds after startup:

Ripple is basically negligible and completely symmetric. Also, the transistor does not even near saturation, with Vce always above 1V.

Why &quot;capacitance multiplier&quot;?
(edited after reading other answers and comments) From the point of view of the load, the capacitance appears multiplied by a factor K. Of course the capacitor does not get physically bigger, nor does its energy storage capacity change. What this circuit does is boosting the current in the load so that the capacitor voltage that is replicated in the output (minus Vbe) does not need to change as much as if the charge had to be supplied by the cap.
To show that, under suitable conditions, the corner frequency is lowered with respect to a passive RC filter, here are two circuits side by side:

(I used an ideal transistor in order not to bother with its parasitic capacitances.)
Now, loading is the key to understand the shift in corner frequency: the simple RC circuit is heavily loaded by RL, so much that the steady state output voltage is about one hundredth of the input voltage (that's about 40 dB attenuation). The time constant is (R1//RL)*C and the corner frequency of this circuit is about 160 Hz.
Now, the boosted circuit does not attenuate that much. The capacitor is loaded with beta + 1 times the value of RL, which for a beta of 100 gives an equivalent load of about 1 kohm; this bring the steady state voltage at half the value of Vin, i.e. an attenuation of 6 dB. The time constant is accordingly increased to (R1//(beta+1)RL)*C and so so the corner frequency drops to about 3 Hz. A factor of more than 50.

In the above Bode plot, the passive circuit's response is plotted in blue, while the capacitance multiplier circuit's response is in brown. From the point of view of the load, it is as if the capacitance has been multiplied by more than fifty times. If we plot the response of a passive circuit with 50+ times the capacitance we get a bode plot that shows the attenuation of the R1-RL divider, but has the same corner frequency as the capacitance multiplier (it's shited 40 dB below).
Why is loading key? Well, the bigger RL is, the less current it requires, the less important is the current boost. The &quot;multiplication&quot; effect gets smaller and smaller as the load gets bigger and bigger. So, for example, if we increase the load to RL = 100k, the cap won't even notice the difference in loading and in fact the Bode plot of the passive circuit and the boosted one will not differ at all, neither in the steady state voltage, nor in the the corner frequency. You also won't notice a difference in absolute ripple reduction between the circuits.

The frequency response of the passive and boosted RC filter for RL = 100 kohm are basically the same
But when the loading is significant, the circuit (even with one resistor) does act as if the capacitance is multiplied. This effect is akin to the widening of the bandwidth allowed by negative feedback. The difference is that instead of introducing attenuation, the transistor circuit removes it.
Reference:
You might want to read the 1985 article from the Designer's Notebook column of Radio Electronics:
&quot;A simple solution to power supply's ripple: A Capacitance Multiplier&quot;
Designer's notebook
Radio Electronics 3, 1985