Let $G=\mathbf{Z}/p \mathbf{Z}$ where $p$ is prime, $X\in G$ be a uniform random variable and $Y\in G^{*}$ be any random variable.
Is it possible to have $Z=X+Y \in G$ with a uniform distribution?
If so, is there any condition on the variable $Y$ that guarantees that $Z$ is uniform?
On
Let $U$ be uniform continuous on $[0,1]$ and $V$ be uniform discrete on $\{0,1,\cdots,m\}$. Then $U+V$ is uniform continuous on $[0,m+1]$.
More generally, assume $U$ to be uniform continuous on $[0,1]$ and $V$ to be another random variable such that $U+V$ is uniform continuous on some other interval. By translation, we can assume that this interval is $[0,L]$ for some $L>0$.
Taking moment-generating functions, we see that: $$\frac{e^t-1}{t}\cdot M_V(t)=\frac{e^{Lt}-1}{Lt}\implies M_V(t)=\frac{e^{Lt}-1}{L(e^t-1)}$$
From here, I can't entirely finish - I know that integer $L$ gives discrete uniform $V$, but I'm not sure for which other values of $L$ it gives a proper MGF.
Guessing that $G^*$ denotes the set of non-zero elements of $G$, and hoping that the question has finally stabilized: If $X$ and $Y$ are independent then $X+Y$ is uniform on $G$. Without assuming independence I don't see how you can expect to say anything about $X+Y$.