Non-trivial solution to $\sum^{n}_{i=1}\sum^{n}_{j=1,j\ne i}(x_{i})^{(x_j)}=(\sum^{n}_{i=1}x_i)^{(\sum^{n}_{i=1}x_i)}$ This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention.
For ease of reading,
$$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[q]}=\sum_{i=1,i\ne p}^nx_i^q. \sum\text{ refers to }\sum_{i=1}^n.$$
Note that $S^q$ is not $S^{[q]}$ and $S_p^q$ is not $S_p^{[q]}$.
Define an equation $A_n$:
$$\sum S_i^{[x_i]}=S^S.$$
For example, $A_3$ is:
$${x_1}^{x_2}+{x_1}^{x_3}+{x_2}^{x_1}+{x_2}^{x_3}+{x_3}^{x_1}+{x_3}^{x_2}=({x_1}+{x_2}+{x_3})^{({x_1}+{x_2}+{x_3})}.$$
Without loss of generality, for every non-negative integer solutions (hereinafter called "solutions") for $A_n$, $x_i\le x_{i+1}$ for every $1\leq i<n$, then there are two distinct non-negetive solutions for $A_3$, one is ${x_1}=0,{x_2}=0,{x_3}=2$, and the other is ${x_1}=0,{x_2}=1,{x_3}=1$.
We call a solution for $A_n$ 'non-trivial' if $x_{n-1}\ne0$. The only known non-trivial solution is ${x_1}=0,{x_2}=1,{x_3}=1$ for $A_3$. The problem is: are there any more non-trivial solutions for $A_n$?
If so, please give an example.
Since this question is difficult enough, I will also recieve answers which give some features about every non-trivial solutions.
Update on 2021-06-26: Claim. For every solution to $A_n$, $$S^{S_n}\le n(n-1).$$
Proof.

*

*Lemma 1. Claim.
$$S_i^{[x_i]}\le(n-1)S_i^{x_i}.$$
Proof. If $x_1=0$, then
$$S_i^{[x_i]}=(n-1)S_i^{x_i}.$$
If $x_1\ne0$, then
$$S_i^{[x_i]}\le S_i^{x_i}\le(n-1)S_i^{x_i}.$$

*For every $1\le i<n$, $x_i\le x_{i+1}$, therefore, for every $1\le i\le n$, $x_i\le x_n$.
And for every $1\le i\le n$, $x_i$ is non-negetive integer, therefore, for every $1\le i\le n$,
$$S_i^{x_i}\le S_i^{x_n}\le S^{x_n}.$$

*Therefore,
$$S^S=\sum S_i^{[x_i]}\le(n-1)\sum S_i^{x_i}\le(n-1)\sum S^{x_n}=n(n-1)S^{x_n}, $$
that is,
$$\frac{S^S}{S^{x_n}}\le n(n-1).$$
Since
$$\frac{S^S}{S^{x_n}}=S^{S-x_n}=S^{S_n},$$
$$S^{S_n}\le n(n-1).$$
And that's what we want.
$\tiny{\text{I've typed for an hour and I finished it finally :)}}$
Update on 2021-07-02: Claim. For every non-trivial solution to $A_n$, $$S^{S_n}\le\frac{n^2-3n+6}2.$$
Proof.

*

*Split $x_i$ to $m$ zeros and $(n-m)$ non-zeros, $0\le m\le n-2$,
$$S^S=\sum S_i^{[x_i]}=\sum_{i=1}^mS_i^{[x_i]}+\sum_{i=m+1}^nS_i^{[x_i]}\le m(n-1)+\sum_{i=m+1}^nS_i^{x_i}\le m(n-1)+(n-m)S^{x_n}.$$

*$$S^{x_n}\ge n-m,$$
$$m(n-1)+(n-m)S^{x_n}\le\frac{m(n-1)}{n-m}\cdot S^{x_n}+(n-m)S^{x_n}=(n-m+\frac{m(n-1)}{n-m})S^{x_n}.$$

*$$d(n-m+\frac{m(n-1)}{n-m})/dx=\frac{n(n-1)}{(n-m)^2}-1,$$
which is always positive for $n-\sqrt{n(n-1)}\le m\le n-2$.
We can see that $n-m+\frac{m(n-1)}{n-m}$ is the greatest when $m=n-2$,
$$(n-m+\frac{m(n-1)}{n-m})S^{x_n}\le2+\frac{(n-2)(n-1)}2\cdot S^{x_n}=\frac{n^2-3n-6}2\cdot S^{x_n}.$$

*$$S^S\le\frac{n^2-3n-6}2\cdot S^{x_n},$$
$$S^{S_n}=\frac{S^S}{S^{x_n}}\le\frac{n^2-3n+6}2.$$
And that's what we want.

 A: Split $x_i$ into $z$ zeroes and a partition of $n$ into $k$ (non-zero) parts, $\lambda_j$. Then your equality can be rewritten as $$z(z-1) + kz + \mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} = n^n$$ The solution with positive $z$ is $$z = \frac{-(k-1) + \sqrt{(k-1)^2 + 4n^n - 4\mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j}}}{2}$$ which by a simple parity argument is an integer iff the square root is an integer. Therefore the question reduces to which partitions into more than one part satisfy $$4n^n + (k-1)^2 - 4\mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} = \square$$
Noting that there are $k(k-1)$ terms in the sum, each of which is a positive integer, and that the non-triviality requirement is that $k > 1$, we see that $\square < 4n^n$, so we rearrange as $$4\mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} - (k-1)^2 = 4n^n -\square$$ where both sides are positive.
Case $n = 2m$
$4n^n = (2n^m)^2$ is a square, so we require $$4\mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} - (k-1)^2 \ge  4n^m - 1$$ to at least reach the next square down.
For $m=1$ we have the known solution; for $m > 1$ the map $x \to x^m$ is superlinear, so a priori it feels hard to satisfy even the relaxed condition $$\mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} \ge n^m$$ In fact, empirically even if we replace $\mathop{\sum\sum}_{i \neq j}$ by $\sum_i \prod_{j \neq i}$ we fall short for $n > 2$.
Case $n = m^2$
$4n^n = (2m^n)^2$, so we get a very similar condition
$$4\mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} - (k-1)^2 \ge 4m^n - 1$$
and again, empirically we fall short even replacing the inner $\sum$ by $\prod$.
Case $n = 2m+1$ is not a square
This is the tricky case. The nearest square is $\lfloor 2n^m \sqrt{n} \rfloor^2$.
If we let $s = \lfloor 2n^m \sqrt{n} \rfloor$, $f = \textrm{fpart}(2n^m \sqrt{n})$, then $4n^n = (s+f)^2$ and we require
$$4\mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} - (k-1)^2 \ge 2sf + f^2$$
but this isn't much use unless we can bound $f$ below. Empirically again the RHS grows exponentially for small $n$, but here we can't easily justify that the gap between the maximum LHS and the RHS won't jump down.
So the strongest claim that I can justify with this approach is that, other than the known example, non-trivial examples will have $n$ odd and not a square.
A: The results are shown in this answer.

*

*The upper bound of $S^{S_n}$.
I proved that $S^{S_n}\le n(n-1)$.
@mathlove commented that $S^{S_n}\le n(n-1)/2$.
I proved that $S^{S_n}\le\frac{n^2-3n-6}2$.

*The parity of $S^S$. @PeterTaylor proved that $S^S$ is odd and not a square.

*Zeros. @mathlove commented that $x_1=x_2=\cdots=x_{\lfloor(2n-1)/3\rfloor}=0$.

