Let $G$ be a Lie group. Let $\phi, \psi \in \pi_n(G)$. Consider $\theta \in \pi_n(G)$ defined by $\theta(x):= \phi(x)\psi(x)\phi(x)^{-1} \in G$, where we use multiplication and inversion in $G$ in the definition of $\theta$. My question is whether $\theta = \psi$ in $\pi_n(G)$.
Note that if $\phi =0 \in \pi_n(G)$, then $\theta = \psi$ in $\pi_n(G)$ does hold. Suppose $\phi': S^{n}\times [0,1] \rightarrow G$ with $\phi'(0)=\phi, \phi'(1) = Id\in G$. Then homotopy between $\theta$ and $\psi$ is $\theta': S^{n} \times [0,1] \rightarrow G$ given by
$\theta'(t) = \phi'(t) \psi \phi'(t)^{-1}$. But I'm not sure whether this holds in general.
Also, if $n=1$ it is true for any $\phi$. In the proof that $\pi_1(G)$ is abelian, we have two maps $g, h: D^1 \rightarrow G$ with $g(0)=g(1)=h(0)=h(1)=Id\in G$. Then the square $D^2 \rightarrow G$ filled by $g(s)h(t)$ has the boundaries [g][h] and [h][g] in $\pi_1(G)$ and has diagonal $g(s)h(s)$, which precisely the map we are looking for. So $[g(s)h(s)] = [g][h] = [h][g]$ and in particular $[h(s)g(s)h(s)^{-1}] = [g]$ as desired. I was a bit confused when I tried to generalize this proof to higher $n$.