Is this a typo in a proof regarding action of a permutation group on a set? Recently, I asked a question about action of a permutation group on a set here. Let me summarize it.

Let $\mathrm{S}_{m}$ be the set of all permutations of $\{1,2,\cdots,m\}$. Then $(\mathrm{S}_{m},\circ)$ is a group where $\circ$ is function composition operation.
Show that $$\mathrm{S}_{m} \times \mathbb{N}^{m} \rightarrow \mathbb{N}^{m}, \quad(\sigma, x) \mapsto \sigma \cdot x := \left(x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(m)}\right)$$ defines an action of $\mathrm{S}_{m}$ on $\mathbb{N}^{m}$.

For $\sigma , \tau \in \mathrm{S}_{m}$ and $x \in \mathbb{N}^{m}$, I try to prove $$\sigma \cdot (\tau \cdot x) = (\sigma \circ \tau) \cdot x$$

In this answer, Wuestenfux presents the following proof:

$$\begin{aligned} \sigma \tau \cdot x &= (x_{(\sigma\tau)^{-1}(1)}, \ldots, x_{(\sigma\tau)^{-1}(n)})\\
&= (x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) \\
&= \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})\\
&=\sigma\cdot(\tau \cdot x) \end{aligned}$$

Here he writes $\sigma \tau$ for $\sigma \circ \tau$.

In his proof, I think $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$ is wrong. Instead, it should be $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)})$.
My reasoning:

*

*In $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)})$, the flow of input-output is $\sigma \longrightarrow \tau \longrightarrow x$, whereas it is $\tau \longrightarrow \sigma \longrightarrow x$ in $\tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$. As such, I feel that it is counter-intuitive to have $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$.


*It's clear that $\sigma \cdot x = x \circ \sigma^{-1}$. Then $$\begin{aligned} \sigma \tau \cdot x &= x \circ (\sigma \circ \tau)^{-1}\\ &= x \circ (\tau^{-1} \circ \sigma^{-1})\\ &= (x \circ \tau^{-1}) \circ \sigma^{-1}\\ &= \sigma \cdot (x \circ \tau^{-1})\\ &= \sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)}) \end{aligned}$$
Please check if my reasoning is correct or not!
 A: I found that the blue argument is not correct in @Wuestenfux's proof. 
$$\begin{aligned} \sigma \tau \cdot x &= (x_{(\sigma\tau)^{-1}(1)}, \ldots, x_{(\sigma\tau)^{-1}(n)})\\
&= \color{blue}{(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)})} \\
&= \color{blue}{\tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})}\\
&= \sigma\cdot(\tau \cdot x) \end{aligned}$$

We have
$$\begin{aligned}
\sigma \cdot (\tau \cdot x) &= \color{blue}{\sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)})} \\
&= \sigma \cdot (d_1, \ldots, d_n) \quad \text{where} \quad d_i = x_{\tau^{-1}(i)} \\
&= (d_{\sigma^{-1} (1)}, \ldots, d_{\sigma^{-1} (n)}) \\
&= (x_{\tau^{-1} (\sigma^{-1}(1))}, \ldots, x_{\tau^{-1} (\sigma^{-1} (n))})\\
&= \color{blue}{(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)})} \end{aligned}$$
It follows that $$(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)})$$ and NOT that $$(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$$
