If you have LED turn signal lamps that don't provide enough load for a standard flasher, you may want to try the circuit below.
I's a simple circuit, using a 12 volt, low current relay, with SPDT or DPDT contacts (Single Pole, Double-Throw, or Double Pole Double-Throw,) a resistor, and a capacitor. It takes advantage of the fact that relays pull in, or energize, at a higher voltage than they drop out, or de-energize. With a typical, 12 volt relay, it might pull in at around 9 to 10 volts, and drop out at about 7 volts.
In the circuit below, note that K1-NC and K1-NO are the Normally Closed and Normally Open K1 relay contacts, respectively.
When you energize the circuit, current flows through the normally closed (NC) contact, through R1, to the relay coil and the capacitor. Initially, there isn't enough voltage across the relay coil to energize the relay. As time goes by, the voltage builds up across the capacitor, and eventually, there's enough voltage (about 10 volts, with a 12V relay) to energize the relay. Energizing the relay opens the normally open contact, allowing the capacitor to discharge through the relay coil. after a while, the voltage drops enough so that the relay becomes de-energized (at about 7 volts.) The normally closed relay contact closes again, and the current flows through the contact & resistor to the relay coil / capacitor, and process starts over again. This charge-discharge cycle will repeat as long as power is applied to the circuit.
Now, while all this is going on, when the relay is not energized, the power to the indicator lamp is on. When the relay IS energized, power to the lamp is off.
(If you use a DPDT relay, you can use the second NC contact to make the current flow to both at the same time.)
I haven't included values for the capacitor C1 and the resistor, R1, as they depend on the relay.
The value of R1 should be about that of the relay coil. If your relay coil is 250 ohms, the resistor should be about 250 ohms. The actual value isn't critical - so long as it's within 20 percent or so. The power rating should be calculated as follows.
P = E * E / R
If you had a 220 ohm resistor, this value would be 12V * 12V / 220 Ohm, or, 0.65 Watts. You should have at a minimum of 50 percent 'extra' power capacity, so, multiplying by 1.5, the value needed is 0.98 watts. The next larger resistor power rating is 1 watt.
Now, for the capacitor. We could go through an involved calculation, but, we would need to know the actual resistor value, the relay pull-in voltage, and drop-out voltage. Probably a better approach is to do a little experimentation, using your actual relay and resistor. Start with a 220 microfarad, 16V (or greater) capacitor. If it flashes too fast, increase the capacitor value. If it flashes too slow, decrease the value. (You can wire two capacitors in parallel, and the capacitance is added together. Don't wire them in series for this circuit, as the result will be much too small.
This circuit can be used with 6 volt batteries, or just about any voltage, for that matter, with the proper selection of relay, resistor, and capacitor values. I've also added a diode in series with R1, and a 120 volt DC relay and used the circuit in industrial applications, to flash a 120 AC volt lamp as a warning indicator.