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?
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 $G^{\mathrm{op}}\to S_A$.