How do you prove mathematically this four card trick?

Question

Shuffle a deck of 4 cards. Ask the participant to remember the bottom card. Move top card to bottom. Flip the first card. Cut (not shuffle) the 4 cards in any way (or not at all). Do a two card flip/swap where the first two cards' positions are swapped and they are also flipped. Flip the first card and move it to the bottom. Move the first card to the bottom. Flip the first card. The "outlier card" that is either flipped or not flipped is the original bottom card. Is there a mathematical theorem/property that can prove or generalize this result? I have checked that it works empirically for 4 cards but wondering if this works only in special cases (for example using only 4 cards with 2 card swap/flips but not in 20 card cases with 6 card swaps/flips). For reference, this youtube video shows the trick: https://www.youtube.com/watch?v=oGEcYkwF6d4&t=272s&ab_channel=NationalMuseumofMathematics

Answer 1

It can be extended to more cards.

It relies on the following fact: If you have a pile of cards that alternate face down and face up, flipping over two adjacent cards as a unit will not affect the face up/face down arrangement. If the pile contains an even number of cards, then you can cut it anywhere and the cards will still alternate, so you can repeatedly cut and flip two cards without affecting that. The only thing that differs is whether it goes DUDUDU.. or UDUDUD... .

The trick first sets up the pile such that the cards alternate except that the chosen card is the wrong way around. With 4 cards this is done by flipping the card that is not adjacent to the chosen card. The cut and flip two procedure will then leave this situation intact, i.e. the cards remain alternating except that the chosen card is the wrong way around. At the end you flip over every other card so that they all face the same way, except that the chosen card is still reversed.