Action of $S_{d}$ in $V^{\otimes d}$ In some books is defined an action of $S_{d}$ in $V^{\otimes d}$ as $\sigma (v_{1}\otimes \cdots \otimes v_{d})=v_{\sigma^{-1}(1)}\otimes \cdots \otimes v_{\sigma^{-1}(d)}$ but then 
$\tau \sigma(v_{1}\otimes \cdots \otimes v_{d})=\tau (v_{\sigma^{-1}(1)}\otimes \cdots \otimes v_{\sigma^{-1}(d)})=v_{\tau^{-1} \sigma^{-1}(1)}\otimes \cdots \otimes v_{\tau^{-1} \sigma^{-1}(d)}=\sigma \tau (v_{1}\otimes \cdots \otimes v_{d})$.....
 A: No, that's not how it works.
To understand how $\tau$ affects the $v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(d)}$, you have to represent it in the same form. So, it would be like $w_1\otimes\cdots\otimes w_d$. One possible choice (and the end result does not depend on the choice you make, so, you can take the one you like) would be $w_1=v_{\sigma^{-1}(1)}$, ..., $w_d=v_{\sigma^{-1}(d)}$. In short, $w_k=v_{\sigma^{-1}(k)}$
Then
$$\tau(\sigma(v_1\otimes\cdots\otimes v_d))=\tau(v_{\sigma^{-1}}\otimes\cdots\otimes v_{\sigma^{-1}(d)})=\tau(w_1\otimes\cdots\otimes w_d)=w_{\tau^{-1}(1)}\otimes\cdots\otimes w_{\tau^{-1}(d)}$$
But, since $w_k=v_{\sigma^{-1}(k)}$, in particular $w_{\tau^{-1}(1)}=v_{\sigma^{-1}(\tau^{-1}(1))}$, $w_{\tau^{-1}(2)}=v_{\sigma^{-1}(\tau^{-1}(2))}$ and so on. Therefore
$$\tau(\sigma(v_1\otimes\cdots\otimes v_d))=v_{\sigma^{-1}(\tau^{-1}(1))}\otimes\cdots\otimes v_{\sigma^{-1}(\tau^{-1}(d))}=v_{(\tau\sigma)^{-1}(1)}\otimes\cdots\otimes v_{(\tau\sigma)^{-1}(d)}=(\tau\sigma)(v_1\otimes\cdots\otimes v_d)$$
