Could you kindly point out what is wrong with my reasoning?

**EDIT: What I have unintendedly proven through my reasoning is that every field can only have one automorphism- the identity mapping. Hope this helps in navigating the mess below**

Let $k\_{1},k\_{2},\dots k\_{n}$ be an arbitrary number of **distinct, non-zero** elements $\in\mathbb{K}$. $\mathbb{K}$ is a field of $0$ characteristic.

Also, $\sigma\_{1},\sigma\_{2},\dots \sigma\_{m}$ are **distinct** automorphisms of $\mathbb{K}$. 

I will show that $m\leq n$. 

Let us assume that $m>n$. We can form the following system of linear equations:

$$\sigma\_{1}(k\_{1})x\_{1}+\sigma\_{2}(k\_{1})x\_{2}+\dots \sigma\_{m}(k\_{1})x\_{m}=0$$ $$\sigma\_{1}(k\_{2})x\_{1}+\sigma\_{2}(k\_{2})x\_{2}+\dots \sigma\_{m}(k\_{2})x\_{m}=0$$ $$\dots$$$$\dots$$$$\sigma\_{1}(k\_{n})x\_{1}+\sigma\_{2}(k\_{n})x\_{2}+\dots \sigma\_{m}(k\_{n})x\_{m}=0$$ 

As $m>n$, we are assured a non-zero solution for $(x\_{1},x\_{2},\dots x\_{m})$. Let it be $(a\_{1}, a\_{2},\dots a\_{m})$, not all zero. 

Then, taking the first linear equation into consideration, $$\sigma\_{1}(k\_{1})a\_{1}+\sigma\_{2}(k\_{1})a\_{2}+\dots \sigma\_{m}(k\_{1})a\_{m}=0, a\_{i}\in\mathbb{K}$$ 

This is not possible (I can give a proof, but it would unnecessarily lengthen the discussion). 

*Hence, on selecting an arbitrary ($n$) number of elements from $\mathbb{K}$, this result shows that the number of automorphisms $\mathbb{K}$ can have $\leq n$. This directly implies that the number of automorphisms $\mathbb{K}$ has is $1$*- the identity automorphism. 

It is quite obvious that this conclusion is wrong. What is erroneous in my thought process though? Thanks in advance!