No, the way they work is quite different. I'll try to explain as simply as I can:
Look at the properties of a (hypothetical) perfect switch - when turned on it has zero resistance between it's contacts so the power loss as current passes through it is (I * I * 0) = zero, when turned off it has infinite resistance so no current flows (0 *0 * infinity) = 0. So a perfect switch would not dissipate any heat at all. Of course in real life the switch would have a small resistance and it would get hot if passing enough current.
In a switch mode regulator (there's a clue in the name!) instead of controlling the output by inserting a resistance in series with the output, it uses the relative time that FULL power or no power is allowed to reach the output. Because it switches almost completely off or completely on, the series pass transistor works like the perfect switch and produces no heat. If you just used the output as it came from the pass transistor you would have a very coarse spikey voltage going from zero to the input voltage which is obviously not what you want. This is where the diode, inductor and output capacitor come into play. You will note that the diode is connected so it normally DOESN'T conduct but there is a reason for it being there. You will also note that the inductor is in series with the output so all the load current has to flow through it.
It works in cycles like this:
1. The oscillator in the 34063 turns the series pass transistor fully on and it shorts the input to the output pin. The diode is reverse biased so it doesn't conduct and the output capacitor starts to charge through the inductor.
2. The oscillator turns the transistor off. The current through the inductor stops and the magnetic field it created collapses. This produces a voltage with reverse polarity to when the transistor was conducting. The diode now conducts and clamps the input end of the inductor close to ground potential. The output end of the inductor is now lifted up by it's stored energy and again charges the output capacitor.
So the transistor is always hard on or hard off, both conditions of low heat loss and the inductor is doing two things, slowing the rising edge of the output voltage when the transitor conducts and filling in the gaps when it's switched off. you can see why the inductor has an optimal value and why it changes with the switching frequency. Basically, the inductor is working as an energy reservoir, taking over from the transistor when the oscillator turns it off. The speed at which the current reverses in the inductor is also the reason why a fast diode is needed. A normal silicon power diode takes quite a long time to stop and start conducting as the polarity across it changes but in this application it has to stop and start conducting at the oscillators speed. Typically the switching is at around 20 - 30 KHz but a normal 1N4001 type of diode can only manage maybe 2 KHz at best.
To actually control the output voltage, the relative time the transistor is turned on compared to it's off time is varied, this limits the energy that can be delivered to the inductor and hence the output voltage. Inside the 34063 the voltage fed back from the divider resistors (which is proportional to the output voltage) is compared to an internal reference voltage and the circuit uses the difference to set the duty cycle of the pass transistor.
I hope that makes sense. I oversimplified it a bit !
Brian.