Let $\tau$ be a field homomorphism unequal to the identity. Show that the induced affinity on $\mathbb{A}^2(K)$ is not a translation or homothety. Show that the induced collinearity on $\mathbb{P}^2(K)$ is not central.
I have no idea how to even start and I wanted to ask for some hints so that I can work out the solution by myself.
To build some intuition I would recommend using a concrete example to think about this. Let $K=\mathbb C$ be the field of complex numbers, and $\tau$ be complex conjugation.
Assume this to be a translation. Then it would uniquely be defined by one point and its image. Pick one point and its image, drive the translation from that, apply it to some other point and find that the result doesn't match what you get from $\tau$.
Same for homothety, except that would be defined by three points and their images.
I'm not sure about the central colinearity terminology in the projective case, but presumably you'd assume a generic projective transformation, defined by four points and their images, and show that thus doesn't map a fifth point the way $\tau$ does either.
The above was for building intuition. The exercise however requires you to show this for arbitrary fields and homomorphisms, not just one concrete example. You can use the fact that $\tau(0)=0$ and $\tau(1)=1$ for any homomorphism. So if you use coordinates only containing $0$ and $1$ as points to define an assumed translation, homothety or projective transformation then the these transformations will all end up being identities. Now you only have to find a single point that doesn't map to itself under $\tau$ to show that the induced colinearity is not the identity and therefore not the same as the assumed operation.