Implicit Function Theorem and order of terms in graph

Question

This is a bit pedantic, but oh well. The Implicit Function Theorem says that if you have a function $\mathbb{R}^{n+k} \to \mathbb{R}^{k}$, then if the differential to $\mathbb{R}^{k}$ is surjective at a point in $f^{-1}(0)$ (if it exists), we can find a function that gives us $k$ coordinates as functions of $n-k$ coordinates, which can be expressed as $$ F(x_{1}, \dots, x_{n-k}, f_{1}(x), \dots, f_{k}(x)) = 0 $$ where $x = (x_{1}, \dots, x_{n-k}) \in \mathbb{R}^{n-k}$. But we can change around the order also; it also makes sense to write $$ F(x_{1}, f_{1}(x), x_{2}, \dots, x_{n-k}, f_{2}(x), \dots, f_{k}(x)) = 0 $$ as well. What allows us to change the order of terms in the graph of of $f$? I know that in the proof, it doesn't really matter which $k$ coordinates you choose because you can do that exact same thing for coordinates in any order, but is there rather than "re-proving" the theorem for any order, is there any simple transformation I can use to deduce the "changed order" Implicit Function Theorem from the "regular" one? Maybe a map like $$ C: \mathbb{R}^{n+k} \to \mathbb{R}^{n+k} $$ that switches around the second and $(n-k+1)$-th entries of a vector in $\mathbb{R}^{n-k}$. How would one express this formally?