I have an LED with no specs, but it does have a resistor. I want to double the voltage
Get a multimeter.
Switch it to 2000 mA range. Connect leads to mA and COM socket.
Wire the meter in series with the (good set of) batteries. Record the current 'i' converting it from mA to amps (A).
Your additional resistor needs to drop 6 V when passing i. We can calculate the required resistor value from Ohms law.


$$ R = \frac {V}{I} = \frac {6}{i} $$.

Pick the nearest standard value.

You might want to check the power rating required.

$$ P = VI = 6i $$

Chose a power rating greater or equal to that value.



As an alternative, buy a second lamp and just wire the two 6 V units in series in your new fitting. You'll have twice as much light for the same current draw on your battery.