How to move from a right semigroup action to a left semigroup action? Let $S$ be a semigroup and $X$ any set. 
Define a left action of $S$ on $X$ to be a map $\sigma: S \times X \rightarrow X$ with the property that $(st)x = s(tx)$, where we define $gx = \sigma(g,x)$ for all $g \in G$, $x \in X$.
Similarly, a right action of $S$ on $X$ is a map $\tau: X \times S \rightarrow X$ with the property $x(st) = (xs)t$, where $xg = \tau(x,g)$ for all $g \in G$, $x \in X$.
Is there a natural bijection between left and right actions of $S$? Does a left action induce in some natural way a right action?
Basically, if I have proven something for every left action of a particular semigroup, is there a shortcut to prove it for every right action as well?
 A: No.  Right actions and left actions of a semigroup can be significantly different.
(I was about to give the same example as Boris.)  
It's actually very common for the left action of a semigroup on itself to be different from the right action.  This is why we have the concept of left and right Cayley graphs for semigroups.
What you do get is that a left action of a semigroup $S$ induces a right action of the 'opposite semigroup' of $S$, which I'll denote $S^{\operatorname{opp}}$.  This is the semigroup with the same underlying set as $S$, but with multiplication reversed, so that in $S^{\operatorname{opp}}$, for any $s,t\in S$ we have $st = t\circ s$, where $\circ$ is the binary operation of $S$. 
In general a semigroup is not necessarily isomorphic to its opposite semigroup, but it is what is called anti-isomorphic.  An anti-isomorphism is a bijective homomorphism $\phi: S\rightarrow T$ such that $\phi(xy) = \phi(y)\phi(x)$ for all $x,y\in S$. 
Note that the reason you never hear of anti-isomorphisms in group theory is that anti-isomorphic groups are isomorphic.  If $G$ and $H$ are groups and $\phi: G\to H$ is an anti-isomorphism, then $\psi: G\to H$ given by $x\mapsto [\phi(x)]^{-1}$ is an isomorphism.  
If you want to prove a result about actions of some class of semigroups, then if the class is closed under anti-isomorphism (which many natural classes of semigroups are), it will suffice to only prove the result for left actions.
A: Left and right actions of semigroups (unlike groups!) can be arranged quite differently. Let $S$ be a semigroups of left zeros (i.e. with identity $ab=a$). Consider its actions on itself by multiplication (i.e. $X=S$). Then for the right action  $X$ consists of  fixed points, but for left action every $a\in S$ maps all $X$ in the single point $a$.
