Nineteenth IMAR Mathematical Competition

The answer is in the affirmative. To define the desired permutation, let $a$ be a quadratic non-residue modulo $p$, let $i{a}_i\equiv a(\mathrm{mod}p)$, $i=1,2,\dots ,p-1$, and let $a_p=p$. Clearly, the $a_i$ form a permutation of $1,2,\dots ,p$; moreover, $a_{p-i}=p-a_i$, $i=1,2,\dots ,p-1$, and, since $a$ is a quadratic non-residue modulo $p$, $a_i\ne i$, $i=1,2,\dots ,p-1$.

To prove $(\ast )$, let first $i,j,k$ be all (strictly) less than $p$. For convenience, write $\equiv$ for congruence modulo $p$. Then $$\begin{array}{rl}ijk((i-j)a_k+(j-k)a_i+(k-i)a_j)& \equiv ij(i-j)a+jk(j-k)a+ki(k-i)a\\ & =-a(i-j)(j-k)(k-i)\ne 0.\end{array}$$ Let now one of $i,j,k$ be equal to $p$. Since the left-hand member of $(\ast )$ is antisymmetric in $i,j,k$, we may and will assume that $k=p$, so $a_k=a_p=p$. Then $$ij((i-j)a_k+(j-k)a_i+(k-i)a_j)\equiv j^{2}a-i^{2}a=a(j-i)(j+i).$$ The latter is non-zero modulo $p$, and $(\ast )$ follows, unless $j=p-i$, in which case $a_j=a_{p-i}=p-a_i$, and $$(i-j)a_k+(j-k)a_i+(k-i)a_j=(2i-p)p-i{a}_i+(p-i)(p-a_i)=p(i-a_i)\ne 0,$$ since $a_i\ne i$. This completes the argument and concludes the proof.