Creating a powered USB hub, trying to drop the voltage from 12v to 5v (AC/DC converter I have outputs 12v)

What you need is an LDO (Low DropOut) regulator. The Micrel MIC2937 has a dropout of only 370 mV at 750 mA out.

You'll have to check the output voltage of the adapter at maximum current. Connect a 16 \$\Omega\$/10 W resistor and measure the voltage. You can combine resistors you have in your box, keeping in mind that P = V\$^2\$/R. The voltage should be higher than 12.4 V.

The MIC29372 (same datasheet) is a version with adjustable output, which may be of help if the output voltage isn't high enough for 12 V out.

edit
If the voltage is too low, and you really want to use the adapter, you can replace the rectifier diodes with Schottky types. That will give you an extra volt output.

edit 2 (regarding the LM317 hli also proposes)
If it's a 12 V adapter it will deliver about 12 V under load. Saying that the unloaded voltage is high enough for the LM317 is meaningless; you do want to use it, don't you? 12 V, even a bit higher is insufficient for the LM317. Don't design for unreliability and use an LDO.

edit 3 (2012-06-14, re your comment) If the adapter is only going to be used only for powering a relay you don't need a regulator. The relay will only show a small load to the 12 V, so it may well be a few volts higher (unloaded transformers output a higher than nominal voltage). This relay is rated at 12 V, but has a maximum voltage 1.7 \$\times\$ that, so that's 20.4 V, which should be safe. Nominal power is 360 mW, and that will be somewhat higher at 15 V for instance (560 mW). If the voltage at 30 mA load is much higher than the 12 V you could still use a regulator. This will also consume some power, but the sum of regulator and relay power would be less than the relay alone.

Also nice about it is Must operate voltage, which is maximum 75 % of rated voltage. That means it will still work if the adapter's voltage would drop a few volts. The Micrel regulator will drop less than 200 mV at 30 mA, if we go for the adjustable, and choose 10 V out, then we will have proper regulation down to 10.2 V in, and the relay will only consume 250 mW instead of 360 mW. So we win twice.

However I'd like a solution that can sustain 4.5V regardless of the current (within the limits above)

If you are happy with a constant voltage of 4.5V then you should consider a low drop out voltage regulator like this: -

It can be setup to deliver up to 4.5V and at 1A to the load and will only need 4.645V (typically) on the input to sustain the output voltage. Note that it needs a minimum load for it work correctly (1mA) but this shouldn't be a problem given your requirements.

This is typical of a series of many voltage regulators with low drop-out voltage - I'm not saying use this one - I am saying be aware of what TI and other folk may be supplying. It's likely you can find one with current limit circuits built-in. You could also apply a bit of current limiting in this device by having a resistor in series with the voltage feed - this should be chosen to develop just enough voltage across it (at the required load current) to sustain the output. Should load current increase, output voltage will drop.