(Update: I don't know whether to scream at this in happiness or disgust.)
The purpose of introducing two more (or in one mind-numbing case, 22 more!) elements into "Rock, paper, scissors" is to reduce the probability of two people having a tie. (Or you could eliminate chance entirely by turning it into a strategy game!) But what if three people play RPS? If there isn't a tie, there is certainly a loop.
In the 5-element version, excluding ties (which you really can't do anything about other than memorise what 25 things do to one another), there is a 60% probability that a rock, paper, scissors, lizard, spock game will have a 'loop', in which no one emerges as a winner. Now, there's a simple way to prevent loops from occuring at all, without a) breaking the structure, and b) destroying the symmetry.
Now, the standard RPSLS game looks like thus: (image credit: Wikipedia)
Every element has two arrows going away from it, and two arrows coming into it. Just make one each pair of arrows 'weak' and the other 'strong'. So there is one weak subduction and one strong, and one weak submission and one strong. But which ones to pick? Simple: The five arrows on the outside of the figure (you can see them going clockwise) are "weak", say, and the five arrows inside (which go anti-clockwise) are "strong". (Or the other way round: doesn't matter.)
How does this stop loops from occuring?
Firstly, what is a loop? It is two arrows in the same direction and one in the opposite, which brings us back to the element that we started with. But two anticlockwise arrows actually need a third anticlockwise one to close the loop, which isn't available! This is because the anti-clockwise ones "jump" over one element, and after two "jumps", you come back to the one you started with only with a third non-jumping arrow in the same direction. Namely, anti-clockwise. Which doesn't exist, because a non-jumping arrow is an outer arrow, which is always clockwise. So there are no loops with two anticlockwise arrows. (eg. Rock --> Scissors --> Lizard isn't a loop.)
Loops do exist with two clockwise and one anticlockwise arrow (eg. Spock --> Scissors --> Paper--> Spock). These loops have two "weak" arrows and one "strong". So there is an element which is weakly subdued but strongly subdues. And that is the winner in that loop.
The symmetry is not destroyed because every element is a part of three loops, and in one it is subdued strongly, and one it is subdued strongly. (The anti-clockwise arrows heading towards and away from it respectively). So it is the strongest loser in one, strongest winner in one, and the weak loser in one. This applies to every element, so excluding ties, we are always guaranteed a winner!