hacking a flash light - what's the resistance required for this group of LEDs?
You can see from the traces on the bottom of the circuit board that all 8 LEDs are in parallel.  There is also a 3 &Omega; resistor in series.

To find the resistor to drop 9 V to 4.5 V you need to know the current.  Since you don't know the exact specs of the LEDs used, the best way to get that is to just measure it.  The resistance to add is then the voltage to drop divided by the current.

For example, let's say the current is 160 mA with 4.5 V applied.  (9 V - 4.5 V)/(160 mA) = 28 &Omega;.

However, this is a crappy way to use a 9 V battery to light 4.5 V LEDs.  A much better way would be a buck converter to make 4.5 V from the 9 V battery.  That has two advantages.  First, it will use the energy in the battery more efficiently.  Second, the brightness won't go down as the battery drains.  It will go out quickly when the battery gets empty.