**Edit 1:** According to the comment of Andreas Cap I revise the integral formula in the question.

**Edit 2:** I understand from the following post that some part of the previos version of my question has negative answer, so I completely revise the question:

http://math.stackexchange.com/questions/1506133/is-there-an-odd-continuous-map-fs2-to-gl2-mathbbc


Let  a compact  topological group  $G$  with invariant  measure $\mu,$ acts on  $S^{2}$ and $\rho$ is  a  $n$-dimensional unitary irreducible   representation of $G$. Let $f:S^{2}\to  M_{n}(\mathbb{C})$ be  a  continuous  function.  is it true to say that:

>There  exist $x_{0}\in S^{2}$  such that the  following integral is  non invertible:

$$ \int_{G}  \rho(g)f(g^{-1}.x_{0})d\mu  $$




As another question:

>Assume that $\phi$ is  a order $n$  homeomorphism on $S^{4}$ and $\lambda \neq 1$ is  a  quaternion with $\lambda^{n}=1$.Assume that $f:S^{4}\to \mathbb{R}^{4} \simeq \mathbb{H}$ is  a  continuous function. Then there is  a point $x_{0}\in S^{4}$  such that $\sum_{i=0}^{n-1} \lambda^{n-i}f(\phi^{i}(x_{0}))=0$.

Note that the particular case $\lambda=-1$  and $\phi=$ The antipodal map is the classical Borsuk-Ulam theorem.

The motivation for the first question is that the statement is true for $1$ dimensional representation.

The motivation for the last part of the question is that the above statement is true if we replace $S^{4}$ and $\mathbb{R}^{4}$  by $S^{2}$ and $\mathbb{R}^{2}\simeq \mathbb{C}$, respectively  and choose $\lambda \in \mathbb{C}$.