First, observe that we clearly need at least
three clues, as we cannot have more than one completely unclued digit without losing uniqueness.
For the types L and T, this is achievable, as seen here

and here

(Both of these basically solve themselves, so I think there is no need to explain a solution path.)
For the Square variant,
we are not so lucky. I'm not sure if there is a slick argument to be found, why three clues cannot suffice but it is also not terribly hard to check this by hand after some preliminary thoughts: Suppose there is a solution with only three clues. These clues cannot all lie in the same band (otherwise the rows/columns of the other band may be swapped). Hence, there must be two clues in diagonally opposing boxes. By swapping rows and columns in a band if necessary, we may therefore assume wlog that two opposing corners are clued. By the symmetry of this setup, we can then reduce to the case that the remaining clue is in one of the five green cells here:

In each of these cases, it is then easy to quickly find a deadly pattern that cannot be resolved.
On the other hand,
four clues
are plenty sufficient, as for example seen here:
