The Mexican Spiral shuffle is a permutation of $52$ (or $n$) items. You can find the order of the permutation, the number of repetitions that make it do nothing, by taking the lowest common multiple of its cycle lengths.
The shuffle is somewhat complex, so I don't think there is a simple formula based on the number of cards $n$. But you can just do the shuffle once on an ordered deck of n cards, examine the result to find its cycles, and then calculate the order from that.
Here is an example. I'll use $9$ cards.
The cards start off as ABCDEFGHI, where A is the top card. A single mexican spiral shuffle affects the cards like this:
ABCDEFGHI
DFHB IGECA
FB HDIGECA
BFHDIGECA
So the end results is that card A went to the location of I, and card I went to the location of E, and so on. We get the following cycles:
A$\rightarrow$I$\rightarrow$E$\rightarrow$G$\rightarrow$F$\rightarrow$B$\rightarrow$A
C$\rightarrow$H$\rightarrow$C
D$\rightarrow$D
These cycles have lengths $6$, $2$, and $1$. The lowest common multiple of these lengths is $6$, so it takes $6$ shuffles for the cards to return to their original order.
If you do this with $52$ cards you'll find that the top card is part of a cycle of length $34$, the second card is in a cycle of length $10$, the fifth card is in a cycle of length $6$, and finally the 12th and 35th cards don't move (i.e. cycles of length $1$). The LCM of $34,10,6,1,1$ is $510$.
Your examples of (2,3,4,5,6) and (2,2,2,3,3) are not hands for the purpose of the question.
(2♦,3♣,4♥,5♠,6♦) and (2♦,2♣,2♠,3♥,3♣) are hands, and are equally likely because the chance of pulling each of the cards involved is equally likely, up to the symmetry of rearranging the order of cards in the hand.
The idea that it's more likely a card will be junk than help to form a useful hand is a good intuition for why some hands are more or less likely, and can form the basis of a different method of calculating the probability of a given hand or class of hand, but is irrelevant in this case.
Best Answer
If you swap the top card with a card chosen randomly with uniform distribution (including the card itself, in which case the "swap" doesn't change anything), then the new top card is any one of the cards with equal probability. Thus every card also has the same probability of not being that card, that is, of being in the subdeck below the top card. Thus the deck will be properly shuffled if the rest of the procedure properly shuffles that subdeck. But the rest of the procedure is precisely the whole procedure applied to that subdeck, so the result follows by induction, the base case being a deck of $1$ card, which is always properly shuffled.
For the second part of the question, as Lopsy wrote, you'd have to say more about what you mean by "randomly swapping cards around".