If two group actions on a set are both transitive, must they be the same?

Question

Suppose we have a group $G$ and a set $S$ and two different group actions $$ \phi:G \to \mathrm{Sym}(S) \qquad \psi:G \to \mathrm{Sym}(S) $$ so that both $\phi$ and $\psi$ are transitive. Can we conclude from this that $\phi = \psi$? If not, what is a counterexample? If not, are there situations in which this does hold, for example, if $S$ is finite?

Answer 1

$G$ acts on itself by left multiplication, $g(x) = gx$, and by right multiplication, $g(x)=xg^{-1}$.

These actions are both transitive, but are not the same except in very special cases -- e.g. by setting $x=1$ we see that it is a necessary condition (which turns out to be sufficient) that every element has order at most $2$.

Every such group is abelian and a vector space over $\mathbb F_2$, and if its dimension is $\ge 2$, Robert Chamberlain's answer gives another set of different group actions.

The only groups where your property can be true are therefore the trivial group and $C_2$ -- and in both these cases it is actually true, though not very interestingly so.

(It is also trivially true if $|S|\le 1$, which is even less interesting -- and vacuously true if there are no transitive actions at all, such as if $|S|$ is nonzero and fails to divide $|G|$).

Answer 2

$\mathbb{Z}/3$ acts on $\{1,2,3\}$ by the actions defined on its generators $1$, $f_1(1)=2,f_1(2)=3, f_1(3)=1$ and $g_1(1)=3, g_1(2)=1, g_1(3)=2$ and these actions are different.

Answer 3

We cannot, for example if $G=C_2\times C_2$ is generated by $x,y$ where $C_2$ is the cyclic group of order $2$ then $G$ acts transitively on the set $\{1,2\}$ in multiple ways - we can have $\phi:x\mapsto (1,2), y\mapsto (1,2)$ and $\psi:x\mapsto (1,2), y\mapsto\mathrm{id}$