I have a doubt about this. We have an affine isometry of an affine space $X$ of dimension 3.
Now, we know it's the composition of some movement (reflection, rotation, etc), with a traslation, and we have it's matrix expression $f$, a matrix of $4\times 4$.
In all the solutions, when they want to calculate the modulus of the movement (how much it moves), they say that the vector of translation must be in $\ker(f-id)^2$.
I don't know why this is.
Suppose $p$ is a point which is subject to translation by a constant vector $v$ under repeated application of your transformation. Then you have $p'-p=v$, but also $p''-p'=v$. Combine them to $p''-p'=p'-p$ eliminating $v$, then rewrite as $(p''-p')-(p'-p)=0$. Now you have
$$(p''-p')-(p'-p)=p''-2p'+p=f^2p-2fp+p=(f^2-2f+id)p=(f-id)^2p$$
So a point $p$ is in the kernel you mentioned iff its difference vector for two subsequent transformation steps remains the same.
Observe that if the above condition holds, then $p,p',p''$ are collinear, with a line in direction $v$. Since affine transformations preserve collinearity, $p',p'',p'''$ must be collinear as well, and so on. The affine transformation, restricted to that line, is defined by two points and their images, so you know that it acts as a pure translation along that line. So $f^{n+1}p=f^np+v$ for all $n\in\mathbb Z$.
Note that I'd interpret that kernel element $p$ as a point, not a vector of translation as your question suggests. To actually treat $p$ as a point, you'd have to find a proper homogeneous representant, i.e. one with a $1$ in its last coordinate. If there is only one such representant, then it is a fixed point of your transformation, and $v=0$. If there is more than one, you have a fixed line as described in the previous paragraph. You can get the vector of translation by computing $v=p'-p=(f-id)p$ for some $p\in\ker(f-id)^2$.