"Suppose the field $F$ is finite. If $f\colon F\to F$ is any bijection, then we can conclude that $\sum_{x\in F}x=\sum_{x\in F}f(x)$. Let $\alpha\in F$ such that $\alpha\ne 0$. Then $x\mapsto \alpha x$ is a bijection. We conclude that $\sum_{x\in F} x=\sum_{x\in F}\alpha x$, i.e. $(1-\alpha)\sum x=0$. If $F$ has more than two elements, we can pick $\alpha\in F\setminus\{0,1\}$ and are done."
1) How can you just assume F to F is a bijection, and 2) What is going on when there is an $\alpha$ picked from F, as seen in the last line, and can you go into more detail about how that proves our point?
Please explain it to someone who's not too comfortable with fields yet.
For any non-zero $\alpha$, $x \mapsto \alpha x$ is a bijection, since we can exhibit an inverse: $y \mapsto \alpha^{-1}y$.
Hence, if $$y = \sum_{x\in F}\alpha x$$ then every $x \in F$ occurs in the sum, since the map above is a bijection. So by reordering the terms in the sum, $$y = \sum_{x\in F}x$$
This gives $\alpha y = y$ for every non-zero $\alpha$, or $$(1-\alpha)y = 0$$so $\alpha = 1$ or $y = 0$.
We want to show that $y = 0$. $F = \mathbb F_2$, we can't conclude this, since the only non-zero $\alpha$ is $1$. But if $F$ has three or more elements, then taking any $\alpha \notin \{0,1\}$ would allow us to conclude that $y = 0$.