Edit: This is a secondary question on how Ralph's solution can be simplified by choosing an appropriate origin in Euclidean space.
Ralph's solution to my original question, in the context of space groups, states that an isometry $(x,q)$ in a space group $G$ with linear part $q\in Q< GL(Z^n)$, must have a translational part x for which
$X_q=\lbrace x\in\mathbb R^n: (q^{1}+\cdots+q^{m})\cdot x\in \mathbb Z^n\rbrace =Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$
Note first that from the composition of isometries we find that
$(t_1,q_1)(t_2,q_2)=(t_1+q_1\cdot t_2,q_1\cdot q_2)$
$\Leftrightarrow X_{q_{1}\cdot q_{2}}=X_{q_{1}}+(q_{1}-1)\cdot X_{q_{2}}$
This means that we must only consider the $X_q$ for the generators of the finite group $Q< GL(Z^n)$ (i.e. the point group).
After a shift of origin in Euclidean space, i.e. an affine transformation $(v,1)$ with $v\in \mathbb R^n$, we can write that
$X_q'=Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)+(q-1)\cdot v$
Since $(q-1)\cdot v\in Col(q-1)$, we can find for every $u\in Col(q-1)$ a vector $v\in \mathbb R^n$ for which $(q-1)\cdot v=-u$. Thus for a proper choice of origin we can write for a generator q
$X_q=\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$
so that $t_q = X_q\ mod\ \mathbb Z^n$ is a rational number with maximal denominator $|Q|$ (which is the maximal possible m). The question is now, can we find one $v\in \mathbb R^n$ so that this simplification can be done for all $X_q$? For this, the column spaces $Col(q-1)$ for generators q of Q should be linear independent. If we call $S$ the generating set of Q, then this can be expressed as
$\forall q_i,q_j\in q,p\in S: Col(q_i-1)\cap Col(q_j-1)=\lbrace Col(q-1)\cap Col(p-1)=\lbrace 0 \rbrace$
Is this true?
