We can modify my solution method (double-blind where one person/group makes the derangement envelopes and proper ratio of numbers for each in a hat, another person/group makes the bijection between people and numbers, and the labeled envelopes contain blank envelopes that contain the next piece of information, so the two people/groups pass the blank envelopes back and forth) to partially satisfy Sandor’s condition that as presents get opened and Santas are revealed one by one, no extra information is reveal. But only up to a point. We can either do it so no one gets any extra information, or so the only people who get information are those who are linked to someone who has been revealed. The latter allows potentially a few more people to open their gifts before everyone has some extra probabilistic envelope, but the former follows more closely the tenet that nobody gain an information advantage. Neither one will let you get all the way down to the end though.
For a particular N, look at the different shapes of graphs of a derangement. For example, with N=8 we have a single 8-cycle, a 6-cycle and a 2-cycle, a 5-cycle and a 3-cycle, two 4-cycles, a 4-cycle and two 2-cycles, two 3-cycles and a 2-cycle, and four 2-cycles, for a total of seven different graphs. If we assign to them probabilities of being drawn as, respectively, A B C D E F and G, then we can try to calculate those probabilities. Obviously A+B+C+D+E+F+G=1. Also, to make everyone equally likely to be your Santa, we must have the odds of being in a 2-cycle at 1/7, so that tells us (B/4)+(E/2)+(F/4)+(G)=(1/7).
After the first person opens a gift and their Santa becomes known, we have four more equations. Namely, for the person who was picked (who now knows whether he/she is in a 2-cycle, but if not then wants to know the odds of being in a 3-cycle), for the person whose recipient was picked (who has learned nothing new), for the person whose recipient’s recipient was picked (and now knows whether he/she is in a 2-cycle, but if not then wants to know the odds of being in a 3-cycle), and for everyone else. These all give linear equations when set equal to the required probability of 1/6 (but 1/7 for the Santa of the person who opened first), but not all of them are linearly independent.
Similarly, after the second person goes, the first two might form a 2-cycle, one of them might be the other’s Santa but not vice-versa, or they might not be directly linked. For each case, all possibilities from among “Being one of the people who went”, “Being the Santa of one of the people who went”, “Being the Santa of the Santa of one of the people who went”, and “Not being directly linked to either person who went” each give a linear equation, though not all of them are necessarily independent.
This can be continued as long as possible, until we run out of free variables. Those variables are just the probabilities of each shape of graph being selected, so for N-8 we have seven of them, meaning we cannot support more than seven independent equations. I haven’t calculated them all out, but I think for N=8 we can only guarantee that nobody gains any extra knowledge through the second person opening a gift. So when the third person goes, someone may (depending on graph shape) get some extra probabilistic information. I think we need N=10 to bump that up so the third person can go without imparting any information.
Also, in my preliminary calculations, it looks like when the first person goes, no matter what N is, only their Santa’s Santa requires an extra equation. That is, the equations for everyone else are all linear combinations of the existing equations. Can anyone prove or disprove that?