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A glass square of side length $2$ is placed on a coordinate plane. The corner at $(-1,1)$ is marked blue. The corner at $(1,1)$ is marked red. We are allowed to reflect the square over the $x$-axis (a vertical flip) or reflect the square over the $y = x$ diagonal (a diagonal flip). What is the shortest sequence of moves (vertical and/or diagonal flips) which culminates with the glass square oriented so that the two colored corners switch places? (List the sequence of moves, and explain why you think it is the shortest).

I'm pretty sure that the shortest way is in three moves: over $y=x$, then over the $x$-axis, then again over $y=x$. But I'm not sure how to show that this is the shortest... Thank you in advance!

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    Maybe prove it can't be done in 1 move but has to be done in an odd number of moves? – 2017-01-18
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    There's only 4 possibilities for combinations of 2 moves, so you can list all of them and show that none work. Alternatively, as @turkeyhundt was suggesting, you can use the fact that an even number of flips is no flip at all (it's a rotation) and you need a flip. – 2017-01-18
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    Why two moves not working - one flip over the x and the second over x=y – 2017-01-21

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