Let $K$ be a field and $\mathcal E$ be a finite dimensional affine space over $K$. Let $f:\mathcal E \rightarrow \mathcal E$ be an affine endomorphism. Assume that there is some point $A\in \mathcal E$ whose orbit under the action of $f$ has finite cardinality $k\in \mathbb N$, where $k$ is not a multiple of the characteristic of the field $K$. Prove that $f$ admits a fixed point.
I have been struggling trying to solve this exercise. Here is what I have found so far.
Let us assume that $k\geq 2$, otherwise there is nothing to do. The vector $v:= \overrightarrow{Af(A)}$ is not zero. We may express our hypothesis in terms of $v$ and of the linear part $\overrightarrow f$ of $f$. We have $$f(A) = A + v$$ $$f^2(A) = f(A+v) = f(A) + \overrightarrow f(v) = A + v + \overrightarrow f(v)$$
Proceeding in a similar fashion and using induction, one may show that for every $i\in \mathbb N$, $$f^i(A) = A + v+ \overrightarrow f(v) + \ldots + \overrightarrow f^{i-1}(v)$$
Using this identity and the fact that $f^k(A)=A$ lies the following
$$v+ \overrightarrow f(v) + \ldots + \overrightarrow f^{i-1}(v)=0$$
Now, any point of $\mathcal E$ can be written under the form $A+w$ for a unique vector $w$. Such a point is fixed by $f$ if and only if $v+\overrightarrow f(w) = w$. Thus, we must show that $-v$ lies in the image of $\overrightarrow f - Id$.
I can't progress any further though. I did notice that if $P$ denotes the polynomial $1 + X + \ldots + X^{k-1}$, then $1$ is not a root of $P$ as the characteristic of $K$ does not divide $k$. If we could prove that this polynomial is the minimal polynomial of $\overrightarrow f$ (or at least that all eingenvalues are roots of $P$), then we may conclude that $\overrightarrow f - id$ is an isomorphism and we would be done.
Let $B=\mathrm{Bar}(A,A^2,\dots,A^k)$ be the barycenter of $A,A^2=f(A),\dots,A^k=f^{k-1}(A)$, where $k$ is chosen such that $A^{k+1}=f^k(A)=A$ and such that none of the $A^i$ with $i\leq k$ are equal to $A$. Then \begin{align*} f(B)&=f(\mathrm{Bar}(A,A^2,\dots,A^k))=\mathrm{Bar}(f(A),f(A^2),\dots,f(A^k))\\ &=\mathrm{Bar}(A^2,A^3,\dots,A^{k+1})\\ &=\mathrm{Bar}(A^2,\dots,A^k,A)\\ &=\mathrm{Bar}(A,A^2,\dots,A^k)\\ &=B \end{align*} is a fixed point.