If I connect OUT to OE through a fast inverter on a LTC6905 at 20MHz, what will the output look like? I need fast pulses into a 90 ohm transmission line.
This should work in a way, but there are some issues that may not work in your system. First, this applies only to the LTC69085-X fixed frequency parts. Second, the LTC6905 will not directly drive a 90 ohm transmission line (assuming its terminated). You will need a buffer/driver of some sort.
In operation, once the OUT goes high, the inverted signal driving the OE input will go low and OUT will then asynchronously go low as fast as the part can achieve. The OUT pulse width would be the propagation delay of the inverter plus the LTC6905-X's OE-to-OUT delay which is not specified or characterized. This OE-to-OUT delay is not well known, but it should be a few ns. Once OUT is low, the OE signal will go high and then OUT will go high on the next OUT rising edge. As long as the propagation delay is much less than a half cycle of the output frequency this should be the next rising edge of the internal clock signal. With a 20MHz output frequency, this should be achievable.
So, to sum up. Your idea should work and you should get a pulse train with a narrow pulse width and a repetition of 20MHz with the caution that we cannot guarantee the pulse width with any accuracy. A half cycle of 20MHz is 25ns and you should be able to get much narrower than this.