If there is a continuous map $h: S^n \to \mathbb{R}^{n+1}-{0}$ such that $x\cdot h(x)=0$ for all $x\in S^n$ then for any continuous map $f:X\to S^n$, $f$ is homotopic to $-f$.
My work:
Given any such continuous $f:X\to S^n$, I need to find a continuous map $F:X\times I \to S^n$ such that $F(x,0)=f(x), F(x,1)=-f(x)$. I tried to construct such a function using the given assumption $f(x)\cdot h(f(x))=0$ for all $x\in X$.
$$F(x,t)=\frac{|f(x)\cdot h(tf(x))|f(x)+|f(x)\cdot h((1-t)f(x))|(-f(x))}{|| \;|f(x)\cdot h(tf(x))|f(x)+|f(x)\cdot h((1-t)f(x))|f(x)\;||}$$.
This way I get the first two conditions, however, I cannot guarantee that the denominator will always be nonzero. I think I need to construct a similar map, however, all of my attempts so far have been unsuccessful. I would greatly appreciate it if anyone can help me solve this problem.
From $h$ it is easy to create a map $g:S^n \to S^n$ such that $x$ is orthogonal to $g(x)$ for every $x\in S^n$. Now, it is sufficient to show that the antipodal map $A:S^n \to S^n:x \mapsto -x$ is homotopic to the identity, since then $f \sim A\circ f=-f$.
To show that $A$ is homotopic to the identity, consider the circles $cos(t)x+sin(t)g(x)$ for every $x \in S^n$ and try to use these circles to define your homotopy $A\sim Id_{S^n}$.