Bloch Functions as an implication of the Crystallographic Restriction Theorem? I'm studying Bloch Functions and it seems to me safe to assume that they are the most general Eigenfunction of a Hamiltionian with the crystal periodicity.
Now the only considerations made in deriving them used the translational symmetry of the lattice. 
Come to think about it, I couldn't find a reason why any other symmetry (rotations, reflections...) shouldn't be represented by the translational one in a crystal lattice. Searching around, this seems to be a defining property of crystal lattices, stated as the Crystallographic Restriction Theorem.
So here the question(s):
Are all symmetries in a crystal lattice reducible to the transitional one?
If so, is this really equivalent to the Crystallographic Restriction Theorem?
 A: Only if your unit cell contains only one atom. 
The translational symmetry is defined by the unit lattice vectors that define exactly which translations leave your lattice the same, and the Bloch functions are based on these unit lattice vectors. 
However, the unit cell itself can contain several different atoms, which is (confusingly, in my honest opinion) called the basis of the lattice. 
Think about it like this: Let's start with a simple quadratic lattice in 2D, with only one type of atom, and one atom in the unit cell. Clearly, then, that lattice is invariant under rotations of 90 degrees. 
Now, change the lattice type so that every atom of the first type has a neighbor, half a unit-length to the right. 
Translationally, nothing has changed, but now suddenly you can't just rotate the thing by 90 degrees! 
That's why there's only 14 different translational symmetries (called Bravais lattices) but 230 different space groups.
A: In general, the rotation operator does not commute with the translation operator.  This means that, in general, the rotation and translation operator do not share a set of eigenvectors.  If we then postulated that we could rewrite one of these operators in terms of the other, we find a contradiction as an operator always commutes with itself.  Thus, the answer to your question would be no.
