Is the natural action of $S_n$ on $\{1, 2, \dots, n\}$ a left or a right action? The symmetric group $S_n$ acts naturally on the set $\{1,2,\dots, n\}$. Is this a left or a right action?
 A: If you think of permutations as functions, then the action is on the left as indicated by Theo.
There is, however, a slight wrinkle when you write your permutations as products of cycles. Some books/people like to perform the multiplication of cycles "left to right", as you are reading the cycles, while others do it right to left. For example, if $\rho=(1,2,3,4)$, $\sigma=(1,2,3)$, and $\tau=(1,4)(2,3)$, and you write $\rho\sigma\tau$, then there are two ways of computing it:
$$(1,2,3,4)(1,2,3)(1,4)(2,3)$$
Working "left-to-right", you'd say: 

$1$ maps to $2$, then $2$ maps to $3$, then $3$ maps to $2$, so $1\mapsto 2$. $2$ maps to $3$, $3$ maps to $1$, $1$ maps to $4$, so $2\mapsto 4$. $3$ maps to $4$, then $4$ maps to $1$, so $3\mapsto 1$; and finally $4$ maps to $1$, $1$ to $3$, and $3$ to $2$, so $4\mapsto 2$. Thus, $\rho\sigma\tau = (1,2,4,3).$

I'm not in my office, so I cannot confirm, but I believe Herstein's "Topics in Algebra" uses this convention.
On the other hand, you could read them right-to-left, as a composition of functions. In that case, you would say:

$1$ maps to $4$, then $4$ is fixed, then $4$ to $1$, so $1$ is fixed. $2$ maps to $3$, then $3$ to $1$, then $1$ to $2$, so $2$ is fixed. $3$ maps to $2$, then $2$ to $3$, then $3$ to $4$, so $3\mapsto 4$. And $4$ maps to $1$, then $1$ to $2$, then $2$ to $3$, so $4\mapsto 3$. Therefore, $\rho\sigma\tau = (3,4)$.

If compute your products of cycles in the former manner, then your action is a right action. If you compute them in the latter, then your action is a left action, just like Theo described it.
A: First of all, there is no essential difference between a left or right action. A left action $\lambda: G \times S \to S$ becomes a right action $\rho: S \times G \to S$ by setting $\rho(s,g) = \lambda(g^{-1},s)$.
Next, it really depends on how you think of $S_{n}$, how you realize it and on your conventions. If you think of $S_{n}$ as bijections $\pi: \mathbf{n} \to \mathbf{n}$ with composition, and if you adhere to the convention that functions take arguments on the right, then the action is on the left: $(\sigma \circ \pi)(k) = \sigma(\pi(k))$.
