Yes, it is possible. The procedure I came up with is really tedious to describe in detail, but I'm attaching a sketch that hopefully should make things clear.

For $k \leq n$ denote by $T_{n,k}$ the following tiling of a $k \times n$ grid. The first tile uses all $k$ squares of the leftmost column, and the leftmost $n-k+1$ squares of the bottom row. The $i$-th tile for $i > 1$ consists of all squares that are incident (vertically, horizontally, or diagonally) to squares of the $(i-1)$-th tile, but have not been used in any other tile.

In particular, $T_{n,n}$ is the tiling where each column is a tile, and $T_{n,1}$ consists of a single tile containing all squares. It's not hard to check that each tile of $T_{n,k}$ consists of exactly $n$ squares. The top left sketch in the attached picture shows $T_{n,k}$ for $n=8$, $k=5$.

I claim that there is a sequence of swaps taking us from $T_{n,k}$ to a tiling where one tile occupies the top row, and the remaining rows form a tiling $T_{n,k-1}$. Applied inductively, this allows us to go from $T_{n,n}$ (all columns) to the tiling where every row is its own tile.

As I mentioned, this sequence is a bit tedious to describe, but I hope that it is clear from the attached picture: in every step, each tile is a path. We swap the green initial piece of one of the paths for the red final piece of the path starting in the top left corner (the tiles stay connected, if we do the swaps in the order indicated by the arrows).

After such a sequence of swaps, the last square of the path starting at the top left corner will be incident to the next path we'd like to perform swaps on. Conversely, the first square of that next path will be incident to the path starting at the top left corner, so we can keep going until one of the tiles occupies the top row. Note that in the $i$-th step we are transforming the $(i+1)$-th path of $T_{n,k}$ into the $i$-th path of $T_{n,k-1}$ on the bottom $k-1$ rows, so the result of our swap sequence is as claimed.