Frobenius on projective variety is not an isomorphism?

Question

This is exercise 1.8 from Arithmetic of Elliptic Curves (Silverman). Part 3 confuses me because isn't $\phi$ the identity, thus an isomorphism?

Let $\mathbb{F}_q$ be a finite field with $q$ elements and let $V \subset \mathbb{P}^n$ be a variety defined over $\mathbb{F}_q$.

1. Prove that the $q$th-power map $\phi = [X_0^q,\ldots,X_n^q]$ is a morphism $\phi : V \rightarrow V$. It is called the Frobenius morphism.
2. Prove that $\phi$ is one-to-one and onto.
3. Prove that $\phi$ is not an isomorphism.
4. Prove that $V(\mathbb{F}_q) = \{P \in V : \phi(P) = P\}$.