Given a vector $x$ in the $n=6$-dimensional Euclidian space $\mathbb{R}^n$, do there exist $n-1$ continuous functions $f_1$ to $f_{n-1}$ such that the matrix $$(x,f_1(x), … ,f_{n-1}(x))$$ is invertible for all $x$ in $\mathbb{R}^n$.
I am aware of the results of
T. Wazewski, Sur les matrices dont les elements sont des fonctions continues. Composito Mathematica, tome 2 (1935), p. 63-68
B Eckmann, Mathematical survey lectures 1943-2004. Springer 2006.
There is no topological obstruction for $n=6$ whereas there is for $n$ odd. I know the solution for $n=2$ and $4$. It is based on permutations with appropriate signs in order to make $f_i(x)$ orthogonal to $x$. This does not work for $n=6$. Hence the question for $n=6$ and even larger $n=8$, …
Restricting $x$ to the unit sphere and projecting all the vectors $f_i(x)$ onto the orthogonal complement of $x$, this is equivalent to asking whether the sphere $S^{n-1}$ is parallelizable (since the $f_i(x)$ will be a basis for the tangent space at $x$ for each $x\in S^{n-1}$). By a famous deep theorem (see, for instance, section 2.3 of these notes by Allen Hatcher), the only parallelizable spheres are $S^0$, $S^1$, $S^3$, and $S^7$. In particular, $S^5$ is not parallelizable, so this is impossible in your case.