TITLE: On existence of boards that be covered by every free tetromino
QUESTION [28 upvotes]: There is a board which can be covered by each of five free tetrominoes:

However, it's not simply-connected (has a hole). I wonder if there is a simply-connected board with the same property.

REPLY [2 votes]: Something like this may work to prove a figure that can be tiled by both squares and skew tetrominoes must have holes. (This builds on some ideas from Yog Urts answer.)
We can always find a cycle of squares where they are connected when they are overlapped by the same tile in the skew tiling. Trough a series of transformations (which I'll describe below), all such cycles can be reduced to the minimum cycle which surrounds only one cell. Because transformations can only reduce the area of the interior by an even amount, it follows the original area of the interior of the inside must have been odd, and therefor cannot be tiled by any tetrominoes.
There are a few "holes" in the argument (maybe somebody can see how they can be removed?):

The cycle exists, but for the transformations to be possible, the cycle of squares needs to be tileable by skew tetrominoes as well (I mean tileable on its own). This needs to be shown.
There may be more transformations than the three I list below.
It needs to be shown the transformations are always possible unless we have the base figure.

Here are the transformations.
If this is the original cycle:

The first transformation is bringing in an edge, that reduces the interior area by 2.

We can apply it again to make the figure look like this:

The second transformation is flipping a corner, that reduces the interior area by 4:

The third transformation is removing a double edge, leaving the interior area the same:

And after a few more transformations we end up at the base figure: