Thanks @Billabob, @Wish Lin, @HonestBook, and @shadowslice e.
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What solutions do you find?
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Based on this insightful observation that (R U)105 = I gives (R U)104 R = U' in 209 turns--which I actually knew is true--I found some shorter solutions with the help of this Order Calculator...

1. The ideal answer I am looking for is if it's possible to turn each layer only once clockwise at a time, and that the opposite layer can not be turned immediately after.
We are looking for an alg to describe U' in terms of U, D, R, L, F, B, i.e., clockwise quarter turns without repetition and...

Indeed! But you read it, and I read the OP. It intrigued me and I spent weeks figuring the algs out. And so, where else to put it? Anyway, for those who find this, it'll be some kind of treasure. I don't mind :)
Although I like identity algs in their own right, they seem to have an application...

Yes, any rotations (where you put the first turn to the end or vice versa) and inversions are valid as well.
Proof:
If for any algs X and Y, X Y = I, then
- Y X = Y X I = Y X (Y Y') = Y (X Y) Y' = (Y I) Y' = Y Y' = I, and
- (X Y)' = I' = I.
EOP.
Some identity (self-solving) algs were...

@xyzzy I was surprised by your answers and I had to think a while whether your answers are actually answering my question.
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Your answer to question 1 does not actually answer the how. I am confused how this could be done quickly. Yes, you could actually inspect all the corners and determine...

I noticed something this week that I've never seen pointed out.
In the 2x2 method, you might end up with the PBL-case where one layer is solved and the other layer is scrambled (opposite corners need to be swapped). But this case requires 11 turns, and that equals the diameter of the 2x2 cube...