Q:

If $G$ acts on $G/H$ and also on $Orb(x)$. Then for all $g \in G$ and for all $aH \in G/H$, we have $\bar{f_x}(g\cdot aH)= g \cdot \bar{f_x}(aH)$

Let $G$ be a group that acts on a set $X$. Let $x \in X$ and $H=Stab(x)$.
1) Let $f_x: G \rightarrow Orb(x)$ defined as $f_x(g)=g\cdot x$. Show that there exists a unique mapping $\bar{f_x}: G/H \rightarrow Orb(x)$ such that $f_x = \bar{f_x} \circ p$ where $p$ is a canonical projection from $G$ to $G/H$. 

I had no issues with this question. Then I am asked to do this:

2)The group $G$ acts on one hand on $G/H$ and on the other hand on $Orb(x)$. Show that $\forall g \in G$ and for all $aH \in G/H$ we have $$\bar{f_x}(g\cdot aH)= g \cdot \bar{f_x}(aH)$$ 

And this is my attempt:
As $\forall h \in aH$ we have $f_x(a)=f_x(h)$, we have $\bar{f_x}(g \cdot aH)= f_x(g \cdot a) = (g \cdot a)\cdot x = g \cdot (a\cdot x)=g\cdot \bar{f_x}(aH)$.
Is my attempt correct?

A:

I think you're right on the money, but I would do it as follows (explaining all), to avoid confusions with notation.
If I understood correctly, we have
$$f_x(a):=ax=ahx\;,\;\;\forall h\in H\implies \overline f_x(aH)=f_x(a)$$
and this would be, apparently, connected with (1) . From here, we obtain:
$$\overline f_x(g\cdot aH):=\overline f_x(gaH)=f_x(ga)\;,\;\;\text{whereas}\;\;g\overline f_x(aH)=gf_x(a)$$
and this is trivial from the definition, both of $\;f_x\;$ and of action of a group:
$$f_x(ga):=(ga)x=g(ax)=gf_x(a)$$

