Let $p$ be a sufficiently large prime number. Then we can define uniquely a function $\varepsilon: \{1,\ldots,\frac{1}{2}(p-1)\} \to \{-1,+1\}$ and a permutation $\sigma$ of $\{1,\ldots,\frac{1}{2}(p-1)\}$ such that $\varepsilon(x)\sigma(x)$ is the multiplicative inverse of $x$ modulo $p$, for each $x=1,\ldots,\frac{1}{2}(p-1)$.
Question. Is it true that, "on average", $\varepsilon$ takes half of the times value $+1$ and half of the times value $-1$?