In the work of my thesis I came up with a problem that is elementary, but I can't figure out its proof.
Let $p$ be an odd prime, let $(\mathbb{Z}/p^n\mathbb{Z})^\times$ denote the multiplicative group of the invertible elements of $\mathbb{Z}/p^n\mathbb{Z}$. Consider $0<a<p^{n-1}$ an integer such that $p\nmid a$, hence we can consider $a$ as an element of $(\mathbb{Z}/p^n\mathbb{Z})^\times$. Let now $\chi:(\mathbb{Z}/p^n\mathbb{Z})^\times\to\mathbb{C}^\times$ be a Dirichlet character (i.e. a group homomorphism). We know that we can extend the domain of $\chi$ to $\mathbb{Z}$.
1) Is it true that $\sum_{j=0}^{p-1}\chi(a+jp)=0$?
2) How can I prove it?
Not necessarily. We could easily have a case that $\chi$ is trivial on the subgroup $$H=\{\overline{m}\in\Bbb{Z}/p^n\Bbb{Z}\mid m\equiv1\pmod p\}\le(\Bbb{Z}/p^n\Bbb{Z})^\times.$$ In other words $\chi(m)=1$ for all $m\in H$.
If that is the case then every term in your sum is equal to $\chi(a)$, and the sum is equal to $p\chi(a)$. This is because $a+jp\in aH$ for all $j$.
In the comments a more interesting variant of the question emerged. The extra assumption that $\chi$ should be a primitive character has an effect, but won't allow us to conclude that the sum would vanish.
For example consider the case $p=3, n=3$. We know that $2$ is a generator of $\Bbb{Z}/27\Bbb{Z}^\times$. We easily see that $2^2\equiv 4$ and hence $2^{-2}\equiv 7$ as $4$ and $7$ are each others modular inverses. So if $\chi$ is defined by $\chi(2^m)=\zeta^m$ where $\zeta$ is a primitive complex root of unity of order $18$, then with $a=1$ your character sum becomes $$ \sum_{j=0}^{p-1}\chi(1+3j)=\chi(1)+\chi(4)+\chi(7)=1+\zeta^2+\overline{\zeta^2}. $$ This is most emphatically not zero for most choices of $\zeta$.
On the other hand, if $n=2$ then the sum vanishes for primtive characters.
Let, again, $$H=\{\overline{m}\in\Bbb{Z}/p^2\Bbb{Z}\mid m\equiv1\pmod p\}\le(\Bbb{Z}/p^2\Bbb{Z})^\times$$ be the prescribed subgroup. As $\chi$ is assumed primitive, its restriction to $H$ is not the trivial character. Therefore $$ S(H,\chi)=\sum_{x\in H}\chi(x)=0 $$ by the usual argument.
What is special about the case $n=2$ is the following. As $j$ ranges from $0$ to $p-1$ the residue class $a+jp$ ranges over the coset $aH$. Therefore uder these assumptions $$\sum_{j=0}^{p-1}\chi(a+jp)=\sum_{x\in H}\chi(ax)=\chi(a)S(H,\chi)=0.$$
When $n>2$ the preceding comment generalizes to (again assuming a primitive character) $$ \sum_{j=0}^{p^{n-1}-1}\chi(a+jp)=0 $$ as well as $$ \sum_{j=0}^{p-1}\chi(a+jp^{n-1})=0. $$ Or even for some $k$, $0<k<n$: $$ \sum_{j=0}^{p^k-1}\chi(a+jp^{n-k})=0. $$ In the latter two cases we are summing over a coset of a smaller subgroup.