Consider the two extreme cases, of wide v/s narrow:

x = [* * *]y = [#]    [#]    [#]

• Consider x = [* * *]. It's very wide/fat, so it doesn't like much exercise, which is why it's columns stabilizer $C_x =\{ e\}$ is trivial. Thus, the action $A_x \equiv id$.

• Consider y = [*][*][*]. It's very slim, and exercises quite a bit. So it's column stabilizer is $S_3$, and its action $A_y \equiv \dots$ has a lot of exercise.

• Anyone can participate in $x$'s exercise regime. In particular, $A_x(y) = id(y) = y$ since $y$ doesn't tire out from the exercise regime of $x$.
• On the other side, it's hard to take part in $y$'s exercise regime and not get TODOed out. If we consider $A_y(x)$, we're going to get zero because by tableaux, there are swaps in $A_y$ that leave $x$ invariant, which causes sign cancellations. But intuitively, $A_y(x)$ is asking $x$to participate in $y$'s exercise regmine, which it's not strong enough to do, and so it dies.

• In general, if $\lambda \triangleright \mu$, then $\lambda$ is wider/fatter than $\mu$. Thus we will have $A_\mu(\lambda) = 0$ since $A_\mu$ is a harder exercise regime that has more permutations.

• Extend this to arrive at specht module morphism: If we have a non-zero morphism $\phi: S^\lambda \rightarrow S^\mu$ then $\lambda \rightarrow \mu$ [Check this?? Unsure ]