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 $\lambda \in \mathbb{C}$.