Jannis Ahlers (Long Bay Primary) found 8 transformations:
"The answer is 8. I found this by finding all the possible positions the shape could end in by only using R, S and there inverses."

The 8 possible transformations are:

I

S

S^{2}

S^{3} = S^{- 1}

R

R S = S^{- 1} R

R S^{2} = S^{2} R

R S^{3} = S R

There are eight transformations made up only of $R$ , $S$ and their inverses. Neat way to see this: draw the eight that you think exist, then note that applying $R$ or $S$ to any of them gives another of them, so we can't `escape' from these eight. The simplest expressions for the eight are:

$I$ , $S$ , $S^{2}$ , $S^{3} = S^{- 1}$ , $R$ , $R S = S^{- 1} R$ , $R S^{2} = S^{2} R$ , $R S^{3} = S R$ .

Notice that $R S R^{- 1} = S^{- 1}$ . (Of course, $R^{- 1} = R$ , so $R S R = S^{- 1}$ , and this can also be written as $S R = R S^{- 1}$ .)

So the two expressions simplified are:

$S S R S R^{- 1} S R S R^{- 1} = S S (R S R^{- 1}) S (R S R^{- 1}) = S S S^{- 1} S S^{- 1} = S$

and

$S^{- 1} R R S R S R R^{- 1} S R^{- 1} = S^{- 1} (R R) S R S (R R^{- 1}) S R^{- 1} = S^{- 1} S R S S R^{- 1} = (S^{- 1} S) R S S R^{- 1} = R S (S R) = R S R S^{- 1} = (R S R) S^{- 1} = S^{- 1} S^{- 1} = S^{- 2} = S^{2}$ .