There are lots of LED blinker circuits around, most driven by a 555 or a transistor-based astable multivibrator. But there are enormously simpler options.
In some sense the simplest LED blinker is two red LEDs in antiseries, in parallel with a capacitor and resistor, in series with another resistor, powered by a voltage around 9–48 V. That is, 9V-R1-(->|-|<- || R2-C)-GND. When the reverse-biased LED goes into avalanche discharge at Vbr, around 5 V, the capacitor discharges down to the minimal voltage necessary to sustain avalanche conduction in the backward-biased LED, after subtracting the forward voltage drop of the forward-biased LED, and with time constant R₂C, with a pulse energy ½Vbr²C, lighting the forward-biased LED. When the current drops too low to sustain avalanche conduction, the reverse-biased LED begins to block again, and the capacitor recharges toward the power supply voltage with time constant (R₁ + R₂)C.
(Most reverse-biased PN junctions have that kind of bistable avalanche behavior, and it’s an annoying source of noise when using avalanche diodes as voltage references, but not all. Note that the plots in Infineon’s appnote linked above do not exhibit this bistability; it is the source of MOS latchup. You could probably force such recalcitrant circuits into oscillation with sufficient series inductance: 9V-R1-L-(->|-|<- || R2-C)-GND.)
This is very similar to the basic neon lamp flasher, except for the R2 current-limiting resistor (which may not be necessary!), the lower voltage, the potentially much higher speed, and the complication that the LED that provides the circuit’s bistability doesn’t light (avalanche-mode LEDs do emit, but at two orders of magnitude lower radiative efficiency), so a second LED is needed.
In theory you could make R1 small enough that the LED would just stay lit until the avalanching LED burned out, but that seems unlikely in practice. And in theory you could leave out R2 and allow the LED to set its own current, since the energy of the pulse will be limited to the energy stored in the capacitor, but even arbitrarily short high currents can damage semiconductors through non-thermal damage mechanisms, so it might be better to leave it in.
Let’s calculate some values so that this circuit is likely to flash visibly. The current through the LEDs should probably be around 20 mA to be brightly visible without much risk of damage, and I think the reverse voltage drop of an avalanching LED is pretty small, maybe around 1 volt. The forward voltage drop of the illuminated LED should be around 1.6 V, so we have about 5 - 1 - 1.6 = 2.4 volts across R2, which means something like 100Ω is appropriate, which would give us 24 mA. Let’s shoot for about 1 Hz overall repetition rate. I suspect that even 1% duty cycle (10 ms) would be bright enough to be visible, but let’s shoot for 50 ms (5%) to be safe.
At this point I realize I don’t have any idea how much current is necessary to maintain avalanche conduction. If I pretend that the avalanche sustaining voltage and the LED forward voltage drop are constant with respect to current, and any amount of current is sufficient to sustain the avalanche, the result is that the capacitor voltage asymptotically approaches the 1 + 1.6 = 2.6 V and never reaches it, and the circuit never turns off, so this is not a useful approximation. I’m sadder but no wiser after reading a couple of application notes about avalanche breakdown.
So, just guessing, maybe you need four RC time constants (54× lower current) to turn off. 50 ms ÷ 100Ω = 500 μF, so a 470 μF electrolytic ought to work. Then we just need to set R₁ to get an off-time of around a second.
The off-time is determined by the time to charge from the avalanche cutoff voltage (plus the forward-biased LED’s voltage drop) up to the breakdown voltage on its way up to the source voltage: from 2.6 V up to 5.0 V out of 9.0 V, which is to say, the remaining voltage should drop from 6.4 V to 4.0 V, a factor of 0.625, about √e, so half a time constant. Thus a time constant of about two seconds would be right, which works out to about 4.3 kΩ, so a 4.7kΩ resistor ought to work.
The visual brightness of the LED should mostly depend on how much charge goes through it, not how long it takes. If we were going to discharge 2.4 volts out of a 470 μF electrolytic at a constant 20 mA, it would take about 56 ms, which is plenty long enough to see the LED flash. So I think probably the LED flashing will be visible. The energy of each flash is about 2.3 mJ, about half dissipated in the resistor, with the other half split almost equally between the two LEDs, so it’s unlikely that the LEDs will be damaged by heating.
The reverse-biased capacitance of the LED is down in the picofarads, so the stored energy is in the dozens of picojoules, insufficient to damage it much.
So the final circuit is:
9V-4k7-(->|[red LED]-|<[red LED]-||100Ω-470μF)-GND
Such a circuit is sensitive to every electronically relevant aspect of its environment: temperature, voltage, light, EMI, and radioactivity. Its off-time is inversely proportional to the difference between the input voltage and the turn-off voltage, though its on-time varies little with that. The avalanche breakdown voltage increases with temperature (Infineon’s appnote above says that in silicon MOS it varies about 5% per 100°), so its both its off-time and its on-time would increase with temperature; but the threshold voltage of the forward-biased diode also changes with temperature, and I think that voltage decreases, which may have a larger effect on the on-time (though it isn’t immediately obvious to me which way). Both diodes are photosensitive, so the circuit can be triggered with a flash of light, and its frequency will vary with the ambient light level. EMI can also advance the timing of the oscillator, so under some circumstances it can phase-lock to weak signals coupled in capacitively or inductively; but even in the absence of EMI, avalanche breakdown is somewhat random, perhaps because it detects radioactive decay as well.
An interesting question is whether this sort of behavior is useful for anything. Of course, if you can mostly isolate the circuit from variations in voltage, light, EMI, and radioactivity, you can use it to measure temperature; if you can mostly isolate it from variations in temperature, voltage, EMI, and radioactivity, you can use it to measure light; etc. But digital logic seems potentially more interesting.
In Implementation and applications of low-voltage Marx generators with solid-state avalanche breakdown? I’ve argued that such triggerable oscillators can be used as clocked logic elements. With ordinary LEDs you ought to be able to get up into the MHz, but with avalanche diodes built on an IC, tens of GHz ought to be attainable. Large resistances or capacitances on chip require large areas, but by running with a supply rail closer to the breakdown voltage and increasing the duty cycle from 5% up to maybe 20%, you ought to be able to use equal or nearly equal resistances for R1 and R2, so neither needs to be particularly large. On chip it might be reasonable to use C ≈ 20 fF, R1 ≈ R2 ≈ 2kΩ, with a frequency around 10 GHz.
Why would you use this kind of dynamic logic instead of regular CMOS? As I understand it, a regular CMOS flip-flop needs 6 (or 8) transistors, and a CMOS NAND gate requires 4. Such an oscillator requires a capacitor, two diode-connected transistors, and two resistors, so it might permit higher density than regular CMOS, but be less power-hungry than four-phase logic (see Snap logic, revisited, and four-phase logic). (But really that's not attacking the crucial aspect of CMOS density, which is routing.)