Suppose a continuous map $f : S^2 \rightarrow S^2$ verifies :
$$ f(x) - f(-x) = 2 \left(f(x),x\right)x $$
where $\left(x,y\right)$ is the scalar product in $R^3$. An equivalent way of expressing $f$ could be :
$$ f(x) = X(x) + u(x)x $$
where $X$ is a tangent vector field of $S^2$ and $u$ a real-valued function with :
$$ X(-x)=X(x) \hspace{15mm} u(-x) = u(x) \hspace{15mm} \left|X\right|^2+u^2 = 1 $$
For instance, the identity and antipodal maps are solutions of degree 1 and -1. My question is : 

Do there exist solutions $f$ of degree $0$ ?

Edit : The answer is no, solutions have odd degree. What about the result for dimension n>2 ?