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 $0$ ?

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 ?