Let $\phi : K^n \to K^n$ be an affinity, such that all lines are parallel to their image under $\phi$. Prove that if $\phi$ has two fixed points, then $\phi$ is the identity mapping.
My attempt:
I denote the two fixed points of $\phi$ by $p$ and $q$. Then I let $x \in K^n$ be arbitrary and aim to prove that $\phi(x) = x$. If $x$ is $p$ or $q$ we are done. If it is not, we can consider the line $[x,p]$ through $x$ and $p$. By our supposition, this line will be parallel to $\phi( [x,p] )= [ \phi(x), \phi(p)] = [ \phi(x), p]$. That is, the two associated linear spaces are the same: $$\operatorname{span}(\, x - p\,) = \operatorname{span}(\, \phi(x) - p\, )\iff x - p = \lambda \,\,(\phi(x) - p),$$ for some scalar $\lambda$. If $\lambda$ were $1$, then I'd be done. But so far I have only used the existence of one fixed point anyway.
I'm not sure how to proceed or if I'm even on the right track so far. A hint would be fantastic.
Note that $[x,p]$ being parallel to $[\phi(x),p]$ gives that the two lines are the same (as $p$ is a common point). On the other hand $[x,q] = [\phi(x), q]$ along the same line of thought. If know $x$, $p$, $q$, do not lie on a common line, we are done, as then $\phi(x)$, $p$, $q$ do not do that either, hence $$ \{x\} = [x,p] \cap [x,q] = [\phi(x), p] \cap [\phi(x), q] = \{\phi(x)\} $$ So $x$ is a fixed point. For the points on the line $[p,q]$, choose a $q' \not\in [p,q]$, which is fixed by the above, and replace $q$ by $q'$ in the argument.