You can easily prove that $ \{2a_{1},2a_{2},\dots.2a_{\varphi(n)}\}$ is a reduced residue system for $ n$ and then $ a_{1}^{1386}+a_{2}^{1386}+\dots+a_{\varphi(n)}^{1386}\equiv (2a_{1})^{1386}+(2a_{2})^{1386}+\dots+(2a_{\varphi(n)})^{1386}(mod n)$ and we get result .

${a_1^{1386}+...+a_{\varphi(n)}^{1386}} \equiv {(2^{1386}-1)a_1^{1386}+...+(2^{1386}-1)a_{\varphi(n)}^{1386} \pmod n}$ so we should prove : $n|a_1 + ... + a_{\varphi(n)}$ .we know $\varphi(n)$ is even and we know $n=a_i+a_{\varphi(n)-i}$ . so done

This problem feels somewhat weird. Notice that $\{2a_1, 2a_2, \cdots, 2a_{\phi(n)}\}$ is also a complete reduced residue system mod $n$ because $n$ is odd. As a result, we have the congruence $$\sum_{i=1}^{\phi(n)} (2a_i)^{1386} = 2^{1386} \sum_{i=1}^{\phi(n)} a_i \equiv (2^{1386} - 1) \sum_{i=1}^{\phi(n)} a_i \pmod n.$$So $n$ divides their difference, which is the result.