Solution to $(1+a_i-a_j)(1+a_j-a_k)(1+a_k-a_i)=(1+a_j-a_i)(1+a_i-a_k)(1+a_k-a_j)$

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Fix a large natural number $n$. Let $a_1,a_2,\dots,a_n$ be $n$ real numbers satisfying

$$ (1+a_i-a_j)(1+a_j-a_k)(1+a_k-a_i)=(1+a_j-a_i)(1+a_i-a_k)(1+a_k-a_j) $$ for any $1\le i,j,k\le n$.

How to find all solutions to the above system of equations? I can see that, obviously, a constant sequence is a solution. But I don't know if there are other solutions.

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Fix $i, j, k$ and expand both sides of the equation. Lots will cancel and you should end up with$$(a_i-a_j)(a_j-a_k)(a_k-a_i)=0$$ so the two expressions are equal if and only if two of the $a$s are equal. You can check that putting two values equal makes the two sides equal very easily. The algebra says there are no other cases.

If any three of the $a_i$ are distinct, they will not give equality. So any sequence made up of at most two values will suffice.

[provided I got my algebra right - but putting everything on the same side I have terms of degree $3$ at most, and setting $a_i=a_j$ gives me zero so I know I have a factor $\pm (a_i-a_j)$ - I'll leave you to check the detail]