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?
['https://math.stackexchange.com/questions/1808887', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/229776/']
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.

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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 $G^{\mathrm{op}}\to S\_A$.
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.

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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 $A$.