Show that the set $\{f_1, f_2, f_3, f_4, f_5, f_6\}$ of functions $\mathbb{R}-\{0, 1\}\rightarrow \mathbb{R}-\{0,1\}$ under composition is isomorphic to $S_3$, where $$f_1(x)= x\\ f_2(x) = 1-x\\f_3(x)=1/x\\f_4(x)=1-1/x\\f_5(x)=1/(1-x)\\f_6(x)=x/(x-1)$$
I am unsure how to start with this. I want to understand how to go about doing this and not just the answer. Any pointers would be a appreciated.
Hint: You need to construct a bijection between your $6$ functions and the $3!=6$ elements of $S_3$. Try writing them down in separate columns and matching them together. Can you spot the identity element? Compose each function with itself until you get the identity function; this will help you figure out its order.
Once you've done that, you should find that $3$ functions have order $2$. Arbitrarily assign these $3$ functions to the three $2$-cycles in $S_3$. To figure out how to assign the remaining $2$ functions of order $3$, pick any two distinct $2$-cycles in $S_3$ and multiply them to obtain a $3$-cycle. Match this $3$-cycle to the function of order $3$ obtained by composing the corresponding functions of order $2$.