We have a group action $\cdot: G\times X\longrightarrow X$,where $G=\mathbb{Z}_{12}\times\mathbb{Z}_3$ and $X=\mathbb{Z}_{12}^{\mathbb{Z}_3}$, which for every $(h,k)\in G$ and $f\in X$ is given by $$((h,k)\cdot f)(a):= h+f(-k+a) $$ where $a\in\mathbb{Z}_3$. The task at hand is to show that if the element $(h,1)$ has a fixed point, then $h\in\{0,4,8 \}$.
My approach is as follows. If $(h,1)$ has a fixed point, then there exists some $f\in X$ such that $f\in\text{Fix}((h,1))$; that is, $((h,1)\cdot f)(a)=h+f(-1+a)=f(a)$. For any choice of $a$, we want the left hand and right hand sides to be equal, so we have \begin{eqnarray*} h+f(2)=f(0) \\ h+f(0)=f(1) \\ h+f(1)=f(2) \end{eqnarray*} This must hold for some choice of $f$, and $f(a)\in\mathbb{Z}_{12}$ but that's where I get lost. My instructor recommended testing specific choices of $f$ to rule out other possible $h$, such as $h=1$, or $(1,1)$. But since the quantifier here is existence, we obviously cannot just choose an $f$ to prove the problem. My current idea is to treat these as a system of equations. For example, if we plug in the first equation to the second, we get $2h=f(1)-f(2)$. Since the right hand side must be divisible by 2 to remain an integer, I think that this difference should be either 0,2,4,6,8, or 10. From here, still confused how to proceed.
I will appreciate any tips. Thanks