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

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.

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