Extrapolating an abstract algebra proof, arriving upon an incorrect conclusion. 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!
 A: The mistake must lie in the assertion that
$$\sigma_1(k_1)a_1 + \cdots +\sigma_m(k_1)a_m = 0$$
is impossible.

Namely, if we let $k_1 = 1$, it is trivial to see that there will be $a_1\ldots a_m$ making this true: just let $a_m = -(a_1+\cdots + a_{m-1})$. This is in direct contradiction with your observation.
Without your purported proof of the above statement, not much more can be said.
A: It looks to me from the comments that you have based your proof a situation in which $\mathbb K$ is an extension of the ground field $\mathbb F$, and the author is examining the automorphisms of $\mathbb K$ which fix $\mathbb F$ in order to get make a comparison with the degree of the extension.
If you have, then your proof essentially assumes $\mathbb F = \mathbb K$, the degree of the extension is $1$, and the automorphisms you are considering are assumed to fix $\mathbb K$. The only automorphism of $\mathbb K$ which fixes $\mathbb K$ is the identity, and what you have done is to prove a special case of the general theorem.
You have not given enough information to show where the assumption about the fixed field comes in - it would be in the part of the proof that you haven't given.
