Show that if $V\in A(G)$ is such that $T_aV=VT_a$ for all $a\in G$, then $V=L_a$ for some $b\in G$ > Define:$T_a:G\to G,T_a(x)=ax$,$L_a:G\to G,L_a(x)=xa^{-1}$. Show that if $V\in A(G)$ is such tath $T_aV=VT_a$ for all $a\in G$, then $V=L_a$ for some $b\in G$. (Hint: Acting on $e\in G$, find out what $b$ should be.) $A(G)$: the symmetrict group for $G$ I tried this: $\forall x\in G, T_eV(x)=VT_e(x)$ but this only leads to $V(x)=V(x)$. I also tried: $\forall a\in G, T_aV(e)=VT_a(e)$, which leads to $V(a)=aV(e)$. So I'm stucked now. What to do next please?

OK, a simpler solution.

For any $x \in G$, $T_xV(e) = VT_x(e) = V(x)$ implies $V(x) = xV(e)$, so if this holds for all $x \in G$, then $V = L_b$ with $b=T(e)^{-1}$.