My question is a follow up of this question Bijection between the free homotopy classes $[S^{n},X]$ and the orbit space $\pi_n/\pi_1$.
For a topological group $G$, there is a natural action of $\pi_1(G)$ on $\pi_n(G)$. I want to show that this action is trivial i.e. each orbit under this action is a singleton.
For $\gamma:[0,1]\rightarrow G$ s.t. $\gamma(0)=\gamma(1)=g$ and $f:\mathbb{D}^n\rightarrow G$ s.t. $f(\delta \mathbb{D}^n)=g$, I have been able to show that there is a based homotopy between the $\gamma_*f$ and $f.\gamma$ where the first one is the normal action of $\pi_1(G)$ on $\pi_n(G)$, and the second one is between the product $(f.\gamma)(r,\theta)=f(r,\theta)\gamma(r)$ where $(r,\theta)$ is the generalized polar coordinates of $\mathbb{D}^n$ (n-dimensional disk).
I don't know how to proceed further.
Any help is appreciated.
Your "product action" can be written as $(f\cdot \gamma)(x) = f(x) \cdot \gamma(\lVert x \rVert)$. I therefore think the basepoint of $G$ should be the neutral element $e \in G$ in order that $(f\cdot \gamma)(S^{n-1}) = f(S^{n-1})$, i.e. that $f \cdot \gamma$ and $f$ map $S^{n-1}$ to the same basepoint.
Now define $$H : D^n \times I \to G, H(x,t) = f(x) \cdot \gamma(t\lVert x \rVert +1-t). $$ Then all $H_t$ have the property that $H_t(S^{n-1}) = f(S^{n-1})$. Moreover $H_0 = f$ and $H_1 = f \cdot \gamma$. This shows that the action is trivial.