TITLE: Let $f_{i}$ be different automorphisms of field $\mathbb{K}$. Does there exist an $x \in \mathbb{K}$ such that $f_{i}(x)$ are pairwise distinct?
QUESTION [8 upvotes]: Let $f_{1},f_{2},\ldots,f_{n}$ be different automorphisms of field $\mathbb{K}$ . What I want to ask is: 

Does there exist an element $x \in \mathbb{K}$ such that $f_{1}(x),f_{2}(x),\ldots,f_{n}(x)$ are pairwise distinct?

REPLY [7 votes]: This is true in the characteristic $0$ case at least.
For any two automorphisms of $\mathbb{K}$, the set of points where they are equal is a subfield of $\mathbb{K}$. Thus a counterexample to the claim gives us an expression of $\mathbb{K}$ as the union of finitely many proper subfields, namely the $\binom{n}{2}$ fields $\{ x|f_i(x)=f_j(x)\}$. Now if $\mathbb{K}$ has characteristic $0$, then it is a vector space over $\mathbb{Q}$, and these subfields are vector subspaces. But a vector space over an infinite field can never be a finite union of proper subspaces.
The case when $\mathbb{K}$ is finite is easy: any finite field is generated by a single element, and different automorphisms must differ on that element. I don't know if the claim is true for infinite fields of positive characteristic.