Show that giving a right-action of a group $G$ on a set $A$ is the same as giving a left-action of $G^{op}$ on A This is a part of an exercise from "Algebra: Chapter 0" by Paolo Aluffi.
First, I provide the necessary definitions.

An action of a group $G$ on a set $A$ is a set-function
$\rho: G \times A \to A$
such that $\forall g,h \in G \ \ \forall a \in A$ we have
$\rho(e_G, a) = a, \ \ \  \rho(gh,a) = \rho(g, \rho(h,a))$

or, alternatively,

An action of a group $G$ on a set $A$ is a homomorphism
$\sigma: G \to S_A$
where $S_A$ denotes the symmetric group of $A$.

We can call this a left-action.
A right-action would be defined the same way expect for "associativity" axiom:

$\forall a \in A \ \ \ \forall g,h \in G \ \ \ a(gh) = (ag)h$

Next:

An opposite group $G^{op}$ of a group $G$ is a group $(G, \times )$ where $a \times b = ba$.

I know that $G^{op} \cong G$ with $f, f(g) = g^{-1}$ being an isomorphism.
Now, I need to show that giving a right-action of $G$ on a set A is the same as giving a homomorphism $G^{op} \to S_A$, that is, a left-action of $G^{op}$ on $A$.
Any ideas?
 A: What “the same” means is that there exists a bijection between the set of left actions of $G$ on $A$ and the set of right actions of $G^{\mathrm{op}}$ on $A$.
Thus we want to find a map $\lambda\mapsto\lambda^r$ and a map $\rho\mapsto\rho^l$ such that, for a left action $\lambda$, $\lambda^{rl}=\lambda$ and, for a right action, $\rho^{lr}=\rho$.
Given a left action $\lambda$, define $\lambda^r\colon A\times G\to A$ by
$$
\lambda^r(a,g)=\lambda(g^{-1},a)
$$
Verifying that $\lambda^r$ is a right action is easy. The definition of $\rho^l$ is similar:
$$
\rho^l(g,a)=\rho(a,g^{-1})
$$
The check that $\lambda^{rl}=\lambda$ and $\rho^{lr}=\rho$ is trivial.

Doing it with homomorphisms is easy, too. If $\sigma\colon G\to S_A$ is a group homomorphism, then
$$
\sigma^\circ\colon G^{\mathrm{op}}\to S_A
$$
defined by
$$
\sigma^{\circ}(g)=\sigma(g^{-1})
$$
is a group homomorphism.
Finish up the argument.

Associating to a left action $\lambda$ a homomorphism $\bar\lambda\colon G\to S_A$ is easy: $\bar\lambda(g)$ is the map $A\to A$ sending $a$ to $\lambda(g,a)$.
Similarly for associating to a right action a homomorphism $G^{\mathrm{op}}\to S_A$.
A: I think the other answers are a bit misled into what the question is asking. It seems that they are answering the last bullet (the sixth bullet) in that exercise (chapter 2, exercise 9.3) in Aluffi's book.

To answer your question (the fourth bullet in that exercise of Aluffi's book): 
If we have a right-action of $G$ on $A$, we have a map $\rho: G \times A \to A$ satisfying $\rho(1,a)=a$ and $\rho(gh,a)=\rho(h, \rho(g,a))$. Note that since this is a right-action, this $\rho$ behaves a bit differently than a left-action.
So, define $G^\circ \to S_A$ by $g \mapsto \sigma_g$ where $\sigma_g: A \to A$ is defined by $\sigma_g(a)=\rho(g,a)$. Then, by the operation $\bullet$ in $G^\circ$, we have
$$\sigma_{g\bullet h}(a)=\sigma_{hg}(a)=\rho(hg,a)=\rho(g, \rho(h,a))=\sigma_g(\rho(h,a))=\sigma_g\circ \sigma_h(a).$$
So, we have a homomorphism $G^\circ \to S_A$.
Furthermore, define $\tau: G ^\circ \times A \to A$ via $\tau(g,a)=\sigma_g(a)=\rho(g,a)$. Then,
$$\tau(g\bullet h, a)=\tau(hg,a)=\rho(hg,a)=\rho(g, \rho(h,a))=\tau(g, \tau(h,a)).$$
Thus, $\tau$ is a left-action of $G^\circ$ on $A$.
