Preimage of surjective linear function from subgroup to base field has same number of elements

Question

Let $G$ be a subgroup of $GF(p^n)$ where $p$ is a prime.

Let $Z_p$ denote the base field $GF(p)$. Consider the linear function $f: G \rightarrow Z_p$. Let $|f^{-1}(c)|$ denote the number of elements $x \in G$ such that $f(x) = c$. If $f$ is surjective, then for $c, c' \in Z_p$, $$|f^{-1}(c)| = |f^{-1}(c')|.$$

I have seen this statement used many times in proofs but I'm a little unclear about why this is true. Can anyone give a proof or intuition, please?

I think this post solves my problem.

Prove the two preimage sets have the same size under a surjective group homomorphism (for a finite group)