Assuming you have 12V relays, and power from a 12V x 300 mA or more DC plug pack, and these are small relays with one Amp contacts or less. (Not car headlight relays etc).
Have two relays, R1 and R2, each with at least one changeover contact. Changeover contacts have the following connections:
Normally Closed (NC)
Normally Open (NO)
Draw out the circuit described here...
From Vs+ to R1 (C) and R2 (C).
From R2 (NC) to coil + of R1.
From coil - of R1 to Vs-.
From R1 (NO) to R2 coil +.
From R2 coil - to Vs-.
LEDs need a series resistor about 1000 ohms (assuming 12V for Vs supply). Connect them between the spare contacts and V-. That allows for two of them. Others could connect across the coils.
The first relay operates the second relay which releases the first which releases the second which connects the first again - forever.
This will happen fairly fast unless the relays have a built in time delay. You can slow it down with capacitors, assuming these are normal DC relays, maybe 1000 micro Farads connected across each coil. Plus to plus, minus to minus. The coils are not likely to be marked + and -, just a and b perhaps.