With a small 12V DPDT relay, a resistor, and a capacitor, you can 'roll your own,' with almost any flash rate you want... Use a DPDT momentary toggle switch as the turn signal switch, and you're good to go. Refer to the attached schematic:
First - the relay should be chosen based on relay coil resistance - the higher, the better. About the greatest resistance you can easily find is in the neighborhood of 250 Ohm for a 12V relay. The greater the resistance, the smaller value of capacitance you can get away with, (and the less power the circuit uses.)
The Resistor should be about 1/2 the coil resistance, at 2 watts power rating. (If the coil is has a lower resistance, the resistor may need to have a larger power rating... If the coil were 100 ohm, for instance, the resistor would be 47 ohm, but it would need to handle at least 4 watts...
You'll need to experiment with the values for R and C some.
The capacitor will be chosen to tune the flash rate. A smaller capacitor will flash faster; a larger cap will flash slower. I would start with a 1000 uF capacitor, and go from there. I wouldn't get one with a voltage rating less than 25VDC. Note that these capacitors are polarity sensitive - make sure you connect the '-' lead to ground...
The switch is a momentary switch, meaning that when you release it, it returns back to the center (off) position. One half of the switch provides power to the relay flash circuit, the other half selects which LED set (left or right) will be lit by the flash circuit.
When the flash circuit is energized, current flows through R1 and charges C1. As the voltage across CI increases, eventually it gets high enough energize the K1 relay. (If R1 is too large, the voltage can never get high enough across the capacitor to pull in the relay, so, if the flash circuit doesn't start flashing, you'll need to make the relay value smaller...) Normally, a 12V relay coil will energize the relay when its voltage is in the 8 to 9 volt range.
When the relay pulls in (energizes) the capacitor is disconnected from the charge circuit, (as the relay contact Normally Closed contact opens, and the capacitor discharges through the relay coil. Once the voltage across the capacitor drops down to about 6 volts, the relay coil doesn't have enough magnetic field to stay energized, and the relay opens. This closes the Normally CLosed contact again, allowing the capacitor to start charging, and the cycle repeats, until you aren't applying power any more.
The second set of relay contacts simply closes and opens, pulsing the current flow through the LEDs. Note that LED turn signals normally have internal resistors, limiting current flow through them. If you are building your own LED turn signals, make sure you have either an LED driver circuit, or current limiting resistors, else you will burn out the LEDs.
This circuit doesn't care what type of bulbs are used - it will handle filament-type bulbs at the same flash rate as LEDs.
If you do decide to go this route, please post back with the relay coil resistance, resistor size, and capacitor sizes you end up with (at what flash rate.)
