I think you need 2 buddies A and B:
1. switch Emmy with A, Prof with B,
2. switch Emmy with B, Prof with A,
3. switch A and B.
For N "scrambled" people, N is an upper bound.
For the lower bound, it depends in what sequence the people got swiched.
For example if A switched with B and C with D, you can undo the mixup with A<->C, B<->D, A<->D, B<->C.
I think it can alwasy be done with 4 extra buddies. I'll show it by example:
If you keep only people "out of their mind" the general configuration is that of a permutation, i.e. a number of cycles: A has B's mind, B has C's mind ... E has A's mind. Let's write Ab for A has B's mind. For example, we could have the configuration with three cycles:
Ab Bc Cd De Ea | Fg Gh Hi If | Jk Kl Lm Mj
Comes first buddy X (Xx) and switches mind with A, B, C in order, the minds are now arranged as:
Ax Bb Cc Dd Ee | Fa Gg Hh Ii | Jf Kk Ll Mm | Xj
Most people are now OK, except for the first in each loop, and X. The remaining people to fix are:
Ax Fa Jf Xj
You can see the remaining people now form a loop (X dragged one mind over to the next loop). To fix that, a new buddy Y comes in and exchanges minds with X, J, F, A (from the end). The minds now are like this:
Aa Ff Jj Xy Yx
Now we are back in the case of Emmy and Prof. Unfortunately X and Y already have switched minds, so 2 more buddies are necessary to switch these minds back.
This is just an example, but I believe it works regardless of the number of cycles. And nobody in the original group needs to swap minds. One shortcut is when there is a single cycle. Then you can jump directly to Y for the final cycle. But it also works with X.
For 3 people, you still have the upper bound of 3 additional buddies. Or you can see that 3 people are always a cycle, so you can fix it Y and the 2 last buddies.
PS: I didn't see there is a better answer already.