If a group homomorphism $f: \Bbb F \to\Bbb C^*$ has $f(ax)=f(bx)$ then $a=b$ Let $\Bbb F$ be a finite field, and let $f: \Bbb F \to \Bbb C^*$ be a  homomorphism from the additive group of the field that is not identically 1.
Assume that there are values $a, b \in \Bbb F$ such that $f(ax)=f(bx)$ for every $x \in \Bbb F$. does that mean that $a=b$?
In case the field is $\Bbb F_p$ this seems to follow from the fact that every homomorphism is of the form $f_m(x) = e^{2 \pi i m \frac x p}$, so plugging in $x=1$ in $f(ax)=f(bx)$ suffices. What happens in general fields?
 A: There is a famous result in Field Theory/Galois Theory which says that all finite fields are of order $p^n$ where $p$ is a prime and $n$ is a positive integer. 
Basically, your finite field $\mathbb{F}$ is a $\mathbb{F}_p$ vector space and any finite dimensional vector space has a basis.
If you are therefore, only looking at the additive structure,
(i.e. $f(a+b)=f(a)f(b)$ for $,ab\in\mathbb{F}$), you just need to tell which root of unity (i.e. $e^{2\pi ik/p}$ for what value of $k$) your bassis vectors have to map to and that defines all the homomorphism uniquely.
Example: Consider the field of order 4. This can be constructed by $\mathbb{F}_2[X]/(X^2+1)$ if you need a concrete construction.
The elements in this field are $1+(X^2+1), X+(X^2+1), 1+X+(X^2+1)$ and $0+(X^2+1)$. We will call these elements as $1,a,1+a$ and $0$ respectively. $1$ and $a$ is a basis as an $\mathbb{F}_2$ vector space. So you have four maps by defining $1\mapsto\pm1$, $a\mapsto\pm1$. Note here that $1$ is allowed here to map to $-1$ as it is not the additive identity and not the multiplicative identity and we are not considering the multiplicative structure.
P.S. So to answer the original question $f(ax)=f(bx) \forall a,b\in \mathbb{F}_q,f\not\equiv 1$ if $a=b$, the answer is correct and actually doesn't require any of the comments above. But I will leave it for illustration as to what the maps would look like.
If $a-b\neq 0$, then $x\mapsto (a-b)x$ is an injection, hence a bijection(This follows from the fact that $\mathbb{F}$ is an integral domain. This mapping is used to show finite integral domains are fields).
Now there exists $y\in \mathbb{F}$ such that $f(y)\neq 1$. In particular, $y=(a-b)x$ for some $x\in\mathbb{F}$. So $f((a-b)x)\neq 1$ so $f(ax)\neq f(bx)$. 
