Q: Consequence of commutating mapd on group $G$ Let $G$ be a group; consider the mappings $\lambda_g:G\to G$ and $\tau_g:G\to G$, defined for $g\in G$ by $\lambda_g(x)=gx$ and $\tau_g(x)=xg$ for all $x\in G$.  If $\theta$ is a one-to-one mapping of $G$ onto itself such that $\lambda_g\theta=\theta\lambda_g$ for all $g\in G$, prove that $\theta=\tau_h$ for some $h\in G$.
Proof:  For any $g\in G$ we have $\lambda_g\theta=\theta\lambda_g$. So for any $x\in G$ we have $\lambda_g\theta(x)=\theta\lambda_g(x)$ or $g\theta(x)=\theta(gx)$. If we put $x=e$ we see that $\theta(g)=g\theta(e)$ and this relation is true for any $g\in G$. Thus $\theta(g)=\tau_{\theta(e)}(g)$ for any $g\in G$. Therefore, $\theta=\tau_h$ on $G$ where $h=\theta(e)$.
Remark: I guess that this problem is trivial and I saw it's duplicate in forum but I have decided not to look at that. So please do not close my topic. However, I spent approximately 1.5 hours in order to come up with solution. Is this solution correct? 
