Topics in algebra Herstein section 2.9, exercise 3. 
If $\theta$ is a one-to-one mapping of $G$ onto itself such that $\lambda_g\theta=\theta\lambda_g$ for all $g\in G$, then $\theta=\tau_h$ for some $h \in G$.

Here $\lambda_g(x)=gx$ and $\tau_h(x)=xh$. 
I tried a lot this exercise, maybe is very simple, but I can't see it. 
The help will be well recieved. 
 A: Hint: $\theta(e)$ is the only possible candidate for $h$.
(BTW, it is not necessary to assume that $\theta$ is a bijection: it follows from $\lambda_g\theta=\theta\lambda_g$ for all $g$.)
A: I don't know if I'm right but this is how I did it.
Lets consider first $\theta x$ since $\theta$ is a bijection $\theta x = r \in G$. $r = xh$ for some h, since $xG = G$. In the same way $\theta (gx) =d \in G$ and $d = (gx)p$ for some p. Now all left is to prove that $h = p, g(\theta x) = \theta (gx)$ by hypothesis, $gxh = gxd$, therefore $h = d$ and $\theta =\tau _h$ for some h. Q.E.D?
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 3 on p.74 in Herstein's book.
My solution is here:

$h(\lambda_g\theta)=h(\theta\lambda_g)$ holds for all $g\in G$ and for all $h\in G$.
So, $(h\lambda_g)\theta=(h\theta)\lambda_g$ holds for all $g\in G$ and for all $h\in G$.
So, $(gh)\theta=g(h\theta)$ holds for all $g\in G$ and for all $h\in G$.
So, $(h^{-1}h)\theta=h^{-1}(h\theta)$ holds for all $h\in G$.
So, $e\theta=h^{-1}(h\theta)$ holds for all $h\in G$.
So, $e=h^{-1}(h\theta)$ holds for all $h\in G$.
So,
$h=h\theta$ holds for all $h\in G$.
So, $\theta$ must be the identity mapping.
So, $\theta=\tau_e$.

But why did the author write "prove that $\theta=\tau_h$ for some $h\in G$.
Why didn't the author write "prove that $\theta$ is the identity mapping"?

I am very sorry.
My answer was wrong.
$\theta$ is not assumed to be a homomorphism.
So, $e\theta=e$ does not hold in general.

I will use the following hint by lhf in lhf's answer above.

Hint: $\theta(e)$ is the only possible candidate for $h$.

My solution is here:

$h(\lambda_g\theta)=h(\theta\lambda_g)$ holds for all $g\in G$ and for all $h\in G$.
So, $(h\lambda_g)\theta=(h\theta)\lambda_g$ holds for all $g\in G$ and for all $h\in G$.
So, $(gh)\theta=g(h\theta)$ holds for all $g\in G$ and for all $h\in G$.
So, $(ge)\theta=g(e\theta)$ holds for all $g\in G$.
So, $g\theta=g(e\theta)$ holds for all $g\in G$.
So, $g\theta=g\tau_{e\theta}$ holds for all $g\in G$.
So, $\theta=\tau_{e\theta}$ holds.


lhf noted the following fact in lhf's answer above:

(BTW, it is not necessary to assume that $\theta$ is a bijection: it follows from $\lambda_g\theta=\theta\lambda_g$ for all $g$.)

In fact, we didn't use the assumption that $\theta$ is bijective.
And $\tau_{e\theta}$ is bijective.
