Kolmogorov Arnold reperesentation theorem states that (the extended version by Lorentz) every multivariate continuous function $f:[0,1]^n\rightarrow \mathbb{R}$ can be written in the folowing form, which basically says that there is a decomposition using univariate functions:
$$f(x_1,x_2,...,x_n) = \sum_{q=0}^{2n} \Phi\left(\sum_{p=1}^{n}\phi_{q,p}(x_p)\right)$$
such that $\phi_{q,p}:[0,1]\rightarrow \mathbb{R}$ and $\Phi:\mathbb{R}\rightarrow\mathbb{R}$.
In a paper by Manzil Zaheer, et al they proved a special case of the above representation for $f:[0,1]^n\rightarrow \mathbb{R}$ and where $f$ is a permutation invariant continuous function (same output when input variables are shuffled):
$$f(x_1,x_2,...,x_n) = \Phi\left(\sum_{p=1}^{n}\phi(x_p)\right)$$
where $\phi:\mathbb{R}\rightarrow\mathbb{R}^{n+1}$ and $\Phi:\mathbb{R}^{n+1}\rightarrow\mathbb{R}$.
They are also referring to a simplified version of the Kolmogorov Arnold representation:
$$f(x_1,x_2,...,x_n) = \Phi\left(\sum_{p=1}^{n}\lambda_p\phi(x_p)\right)$$
where $\phi:\mathbb{R}\rightarrow\mathbb{R}^{2n+1}$ and $\Phi:\mathbb{R}^{2n+1}\rightarrow\mathbb{R}$. For the special case above $\lambda_p = 1$ for all $p$. I cannot see the equivalency to the original theorem immediately but at least it is more readable.
Now my questions is which additional terms from the general representation form I need to add to the special case such that for $f:[0,1]^{n+1}\rightarrow \mathbb{R}$:
$$f(x_0,x_1,x_2,...,x_n)$$
has a representation for a fixed location $x_0$, but permutation invariant for $x_1,x_2,...,x_n$? Would it be enough to set $\lambda_0$ to be some constant other than 1? Probably not. I think the final representation will be more complicated. If it is going to make things easier I can also work with bivariate functions with the form $\phi(x_0,x_p)$ in the inner sum.