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?
 A: This is just the theorem that the centralizer of the image of the left regular permutation representation of a group is the image of the right regular representation.
Let $X = \langle T_a : a \in G \rangle$ and $Y = \langle L_a : a \in G \rangle$. Then $X,Y$ are both transitive subgroups subgroups of $A(G)$ with trivial stabilizers, and it is easy to check that they centralize each other.
Now, for any set $\Omega$, the centralizer in ${\rm Sym}(\Omega)$  of any transitive subgroup of ${\rm Sym}(\Omega)$ has trivial point stabilizers. (I'll leave that as an exercise.)
But if $Y$ was properly contained in the centralizer $C$ of $A(G)$ of $X$ then, since $Y$ is transitive already, $C$ would have nontrivial point stabilizers, contradicting the result above. (This is clear from the Orbit-Stabilizer Theorem if $G$ is finite, but it is still true when $G$ is infinite.) So $C=Y$.
A: 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 $b=T(e)^{-1}$.
