That's a mighty good permutation algorithm you're using. I'm curious as to where you came across it. (A few years ago I did a fair bit of investigation into permutation generation, and this algorithm was the fastest pure-Perl one I could find by quite a margin. Your code looks remarkably similar to mine!)

It may also be worth looking at the algorithm used in tye's Algorithm::Loops, which is jolly clever. I haven't compared it directly with this one, though if I had to guess, I would guess this one is probably still faster.

PS. I still haven't figured out how you can get away with doing nothing when $copy->[-1] == @$copy. What's the trick?

Update: Okay, I've got it. That's pretty clever. Here's my proof that it works.

Lemma: for every $n ≥ 1, there is some permutation of (1..$n+1) that takes more flips than any permutation of (1..$n).
Proof. Let @a be a permutation of (1..$n) that takes as many flips as possible, say it takes $flip flips; and consider the permutation ($n+1, reverse @a). Clearly this takes ($flip+1) flips.

Now, if $copy->[-1] == @$copy then the last element can never be moved by the flipping, and so this will take the same number of flips as @$copy[0..$#$copy-1]. By the Lemma, this is less than the maximum number of flips.