As Ross Millikan points out, if you must swap the first card with some card strictly below it, then the card that starts out on top has probability zero of winding up on top, so the answer to your (first) question is, no. As Ross also suggests, things are more interesting when you allow the first card (and each subsequent card) to stay put; make the top card swap with the $k$th card with probability $1/52$ for each $k$, $1\le k\le52$, then make the second card swap with the $k$th card with probability $1/51$ for each $k$, $2\le k\le52$, etc. etc. Then indeed all cards have equal probability of ending up in any given position; in fact, more is true, as all 52-factorial possible permutations of the deck are equally likely.
This is most easily seen by proving that it's true not just for a 52-card deck, but for an $n$-card deck, for any positive integer $n$, using induction on $n$. The base case is easy. Then assume it's true for $n$-card decks, and assume you have an $n+1$-card deck. Make the first swap; any card could now be on top, all with equal probability, and the card that's now on top never gets moved after this, so you are now dealing with just the remaining $n$ cards. Now apply the induction hypothesis!