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I think I have gotten a little further. I believe we are on the wrong track just flipping one coin. Instead, make a pile of coins containing the same number of heads up coins that was given. Then flip this pile of coins upside down, and both piles will now contain the same number of heads up coins. Here is an example--
Suppose you start with 5 coins with 3 heads up. Your stacks must start as one of these
3h -- 2t
2h1t -- 1h1t
1h2t -- 2h
After flipping the first stack you will have
3t -- 2t
2t1h -- 1h1t
1t2h -- 2h
It looks like it works for any number of coins as long as you know the initial number of heads up coins. I've done it for 3, 4, 5, and 6 coins with different starting heads up numbers. I'm still trying to put together a mathematical reason for why it works.
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