I am trying to prove that $S_{n}$ acts on $\mathbb{R}_{n}$ with the map $$* : S_{n} \rightarrow \mathbb{R}_{n}, \quad * \left( \sigma, \left( r_{1}, r_{2}, \dots, r_{n} \right) \right) = \left( r_{\sigma_{\left( 1 \right)}} ^{-1}, r_{\sigma_{\left( 2 \right)}} ^{-1}, \dots, r_{\sigma_{\left( n \right)}} ^{-1} \right).$$
However, I realized the map is undefined for any $n$-tuple having one or more coordinates equal to zero, so I suspect it is not an action. Am I right? If so, what would be the correct map?
UPDATE 1: After working this out I think, even if the map is defined as $$* : S_{n} \rightarrow \mathbb{R}_{n}, \quad * \left( \sigma, \left( r_{1}, r_{2}, \dots, r_{n} \right) \right) = \left( r_{\sigma_{\left( 1 \right)} ^{-1}}, r_{\sigma_{\left( 2 \right)} ^{-1}}, \dots, r_{\sigma_{\left( n \right)} ^{-1}} \right),$$
that it is not an action of $S_{n}$ on $\mathbb{R}_{n},$ and the reason for this is precisely the identity $\left( \sigma_{1} \circ \sigma_{2} \right)^{-1} = \sigma_{2} ^{-1} \circ \sigma_{1} ^{-1},$ for $\sigma_{1}, \sigma_{2} \in S_{n}.$ Here's what I did:
\begin{align} * \left( \sigma_{1} \circ \sigma_{2}, \left( x_{1}, \dots, x_{n} \right) \right) & = \left( x_{\left( \sigma_{1} \circ \sigma_{2} \right)^{-1} \left( 1 \right)}, \dots, x_{\left( \sigma_{1} \circ \sigma_{2} \right)^{-1} \left( n \right)} \right) \\ & = \left( x_{\left( \sigma_{2} ^{-1} \circ \sigma_{1} ^{-1} \right) \left( 1 \right)}, \dots, x_{\left( \sigma_{2} ^{-1} \circ \sigma_{1} ^{-1} \right) \left( n \right)} \right) \quad \text{(This is the key step)} \\ & = \left( x_{\sigma_{2} ^{-1} \left( \sigma_{1} ^{-1} \left( 1 \right) \right)}, \dots, x_{\sigma_{2} ^{-1} \left( \sigma_{1} ^{-1} \left( n \right) \right)} \right) \\ & = * \left( \sigma_{2}, \left( x_{\sigma_{1} ^{-1} \left( 1 \right)}, \dots, x_{\sigma_{1} ^{-1} \left( 1 \right)} \right) \right) \\ & = * \left( \sigma_{2}, * \left( \sigma_{1}, \left( x_{1}, \dots, x_{n} \right) \right) \right), \end{align}
but we would like to have $* \left( \sigma_{1} \circ \sigma_{2}, \left( x_{1}, \dots, x_{n} \right) \right) = * \left( \sigma_{1}, * \left( \sigma_{2}, \left( x_{1}, \dots, x_{n} \right) \right) \right)$ instead.
Is this correct?
UPDATE 2: I am really confused now. Looking in the web I found two different notes on group actions. In one, they say the map defined as $$\phi : S_{n} \rightarrow \mathbb{R}_{n}, \quad \phi \left( \sigma, \left( r_{1}, r_{2}, \dots, r_{n} \right) \right) = \left( r_{\sigma_{\left( 1 \right)}}, r_{\sigma_{\left( 2 \right)}}, \dots, r_{\sigma_{\left( n \right)}} \right)$$
is not a group action, while in the other one they say it is. I think, by my own calculations, that $\phi$ is a group action, but I'm not sure now.
Can someone help me with this one?