Let $G$ be a group and define
$$\ell_g : G \to G, \qquad \ell_g(x)=gx,$$
$$r_g : G \to G, \qquad r_g(x)=xg.$$
Let $\phi : G \to G$ be a bijection such that $\ell_g \circ \phi = \phi \circ \ell_g$ for any $g \in G$. How can I go about showing that $\phi = r_h$ for some $h \in G$?
Attempt so far: The conditions on $\phi$ say that for any $g,x \in G$, we have $g \phi(x) = \phi(gx)$. I suppose next I should use the fact that $\phi$ is a bijection to show that $\phi(x) = xh$ for some $h \in G$, but I'm not quite sure how. Any hints?
Here is a hint: letting $e$ be the identity, we see that $$g\phi(e)=\phi(g)$$ and thus $$\phi(e)=g^{-1}\phi(g).$$ Given that you're trying to show that $\phi$ is of the form $\phi(x)=xh$ for some $h\in G$, what would $h$ have to be?