$n=1$ or $n$ is even

For $n=1$, take $a_1=0$. For $n$ is even, take $a_{2i}=n-i$ and $a_{2i-1}=i-1$ for $i=1,2,\cdots ,\dfrac n2$.

$a_1, a_2-a_1, a_3-a_2, \dots ,a_n-a_{n-1}$ are different at $\text{mod n}$. Thus, $\{a_1, a_2-a_1, a_3-a_2, \dots ,a_n-a_{n-1}\}\equiv \{0,1,\cdots ,n-1\} \pmod{n}$. We know that $a_i\not \equiv a_j\pmod{n}$ for $i\neq j$. Hence; $a_2-a_1, a_3-a_2, \dots ,a_n-a_{n-1}\not \equiv 0 \pmod{n}$. So, $a_1\equiv 0 \pmod{n}$. $a_1+(a_2-a_1)+(a_3-a_2)+ \dots +(a_n-a_{n-1})\equiv 0+1+\cdots +(n-1)\equiv \dfrac{(n-1)n}{2}\equiv 0 \pmod{n}$. Hence, $a_n\equiv 0 \pmod{n}$. We have $a_1\equiv a_n\equiv 0 \pmod{n}$. Contradiction.