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 $ia_i \equiv a \pmod p$, $i = 1, 2, \dots, p-1$, 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 $p$, $a_i \neq i$, $i = 1, 2, \dots, p-1$.

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