Peter Walters gave a proof for this fact in his book but I don't understand a few steps.
Proof: Let $A : G \to G$ be a surjective endomorphism on a compact abelian group. Suppose that the trivial character is the only one that satisfies $\gamma A^n = \gamma$ for some $n$. We want to show $A$ is ergodic. Let $f \circ A =f $. Our goal is to prove that $f$ is constant. Expanding the fourier series we have
$f=\sum a_n \gamma_n$.
with $\sum a_n^2 < \infty$.
By invariance $\sum a_n \gamma_n(Ax)= \sum a_n \gamma_n$, so that if $\gamma_n , \gamma_n(Ax), \gamma_n(A^2x), ..., $ are all distinct their coefficients are equal and therefore equal to zero. So if $a_n \neq 0$, $\gamma_n(A^p)=\gamma_n$ for some $p>0$. By assumptions $\gamma_n = 1$ and hence f is constant. Thus $A$ is ergodic.
I don't get why distinctness of the iterates implies that they have equal coefficients or why that implies $a_n \neq 0$ implies $\gamma_n(A^p)=\gamma_n$ for some $p$.
Thanks
Let's use better notation. Write $f=\sum_\gamma a_\gamma\gamma$, where we sum over all characters. Then $f\circ A = \sum_\gamma a_\gamma(\gamma\circ A)$. The coefficient of $\gamma\circ A$ in $f$ is $a_{\gamma\circ A}$, and the coefficient of $\gamma\circ A$ in $f\circ A$ is $a_\gamma$. So we have $a_{\gamma\circ A} = a_\gamma$. Similarly, we have $a_{\gamma\circ A^2} = a_{\gamma \circ A}$. So we have $a_{\gamma} = a_{\gamma\circ A} = a_{\gamma\circ A^2} = a_{\gamma \circ A^3} = \dots$. Now, if $\gamma,\gamma\circ A,\gamma\circ A^2,\dots$ are all distinct, we must have $a_{\gamma} = a_{\gamma\circ A} = a_{\gamma\circ A^2} = a_{\gamma \circ A^3} = \dots = 0$, since $\sum_\gamma |a_\gamma|^2 < \infty$.
Therefore, taking the contrapositive, if $a_\gamma \not = 0$, it can't be that $\gamma,\gamma\circ A,\gamma\circ A^2,\dots$ are all distinct. In other words, $a_\gamma \not = 0$ implies there's some $p > 0$ with $\gamma\circ A^p = \gamma$.