Let $d_i=\dfrac{1}{\binom{n}{k}}\sum_{sym}a_1a_2\dots a_i$ (Is a symmetry without repetitions). Basically the problem says that $d_i=a_i^i \Leftrightarrow \sqrt[i]{d_i}=a_i$. Now, Maclaurin's Theorem says that $d_1\geq \sqrt{d_2} \geq \sqrt[3]{d_3}\geq \dots \geq \sqrt[n]{d_n}$, so we have that $a_1\geq a_2 \geq \dots \geq a_n$. Since $a_1n\geq a_1+a_2+\dots+a_n= a_1n$, so necessarily $a_1=a_2=\dots a_n$ and is clear that it fulfills the equation.

Let $d_i=\dfrac{1}{\binom{n}{k}}\sum_{sym}a_1a_2\dots a_i$ (Is a symmetry without repetitions). Basically the problem says that $d_i=a_i^i \Leftrightarrow \sqrt[i]{d_i}=a_i$. Now, Maclaurin's Theorem says that $d_1\geq \sqrt{d_2} \geq \sqrt[3]{d_3}\geq \dots \geq \sqrt[n]{d_n}$, so we have that $a_1\geq a_2 \geq \dots \geq a_n$. Since $a_1n\geq a_1+a_2+\dots+a_n= a_1n$, so necessarily $a_1=a_2=\dots a_n$ and is clear that it fulfills the equation.

We claim that all the solutions are given by $a_1=a_2=\dots=a_n$. In the case $n=1$ there is nothing to do, so assume $n \ge 2$. Let $k$ be minimal such that $\vert a_k\vert=\max_i \vert a_i\vert$. Comparing coefficients of $x^{n-k}$ we have \[\binom{n}{k}a_k^k=\sum_{\vert I\vert=k} \prod_{i \in I} a_i.\]Now each of the $\binom{n}{k}$ summands on the RHS has absolute value at most $\vert a_k\vert^k$ by assumption. So equality must hold everywhere and also in the triangle inequality whence $\prod_{i \in I} a_i=a_k^k$ for all $I$ with $\vert I\vert=k$. If $a_k=0$, then $a_1=\dots=a_n=0$ and we are done. If $a_k \ne 0$, then all $a_i$ are non-zero. If $k<n$, it is immediate that all $a_i$ are equal by just exchanging one element in $I$. Finally if $k=n$ we have $a_1a_2\dots a_n=a_n^n$ and hence all the $a_i$ have the same absolute value, contradicting minimality of $k$. Done.