On the opposite group When the group $G$ acts on the set $S$, e.g. $G \colon= {\mathrm{GL}}_2({\Bbb C})$, $S \colon= {\Bbb C}^{\oplus 2}$, $G$ have the right to act on the left or on the right. Both actions can be naturally identified as follows$\colon$
$\sigma \circ \tau (s) = (s)\tau \circ \sigma$,
where both sides means $\tau$ acts on $s \in S$ first and then $\sigma$ acts on $\tau(s) ~(= \!(s)\tau)$ next.
That is, the equality
\begin{equation*}
\sigma \circ \tau = \tau \circ \sigma \quad \cdots\cdots \quad (\lozenge)
\end{equation*}
automatically makes sense under the sequence of actions (or time).
Or when one starts from the left action, the set in the R.H.S. in $(\lozenge)$ can be put, by the more sophisticated terminology, the opposite group $G^{\mathrm{op}}$, for which the unique isomorphism $G \cong G^{\mathrm{op}}$ is given by $g \mapsto g^{-1}$.
However, the identification in $(\lozenge)$ is accomplished by switching the multiplication between two elements $\sigma$ and $\tau$ according to the sequence of actions on the left or right (on $S$).
Q. Is it correct that I never care about $G^{\mathrm{op}}$ but always automatically identify $G$ and $G^{\mathrm{op}}$ as in $(\lozenge)$ ignoring the group isomorphism between $G$ and $G^{\mathrm{op}}$?
 A: It sounds like you are saying if you start with a left action of $G$ on $S$ (with $g$ acting on $s$ written "$gs$"), that you want to define a "right action" by just declaring that "$sg$" is an alternate notation for $gs$. This is what I assumed when you said $\sigma \tau (s) = (s) \tau \sigma$ both mean do $\tau$ first, then $\sigma$.
Here is the problem with this: the right-hand side, $s \tau \sigma$ is not really consistent. You get different results depending on if you intepret it as meaning $(s\tau)\sigma$ or $s(\tau\sigma)$ (but they need to agree to be a right action).
I will show by example. Consider the example you give, of $G = \mathrm{GL}_2(\Bbb{C})$ acting on $S=\Bbb{C}^2$. Consider
$$ g = \begin{pmatrix} 0&1\\1&0 \end{pmatrix} \hspace{0.5cm} \text{and} \hspace{0.5cm} v=\begin{pmatrix}x\\y\end{pmatrix}$$
Then of course $gv = \begin{pmatrix}y\\x\end{pmatrix}$. So if we define our right action like you suggest, then $vg = \begin{pmatrix}y\\x\end{pmatrix}$.
But $g$ can be factored as $g=g_2g_1$, where
$$ g_2 = \begin{pmatrix}0&-1\\1&0\end{pmatrix} \hspace{0.5cm} \text{and} \hspace{0.5cm} g_1 = \begin{pmatrix} 1&0\\0&-1 \end{pmatrix} $$
So in the usual left action, multiplication by $g$ is the same as first multiplying by $g_1$, and then by $g_2$. So your suggestion is that if we write things on the right, we should have $g_2g_1v = vg_1g_2$.
This is where the problem occurs: do we interpret the expression $vg_1g_2$ as meaning $(vg_1)g_2$ or as $v(g_1g_2)$? To be a valid right action, it shouldn't matter and both should be the same. The first interpretation is $(vg_1)g_2$, meaning act first by $g_1$, and then by $g_2$. This agrees with the calculation above (and the left action), and we get $\begin{pmatrix}y\\x\end{pmatrix}$. The other intrepretation is $v(g_1g_2)$, which means first multiply $g_1g_2$ in the group, and then act by that element. But $g_1g_2 = \begin{pmatrix}0&-1\\-1&0\end{pmatrix}$, which is not the same as $g$, and we get $\begin{pmatrix}-y\\-x\end{pmatrix}$ as the result.
