Why does any non-trivial permutation of ${\rm Aut}(S(5,6,12))$ at least alter $8$ symbols?

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In my study on Mathieu Groups I came across the following article. R.G. Stanton states on the second page:

... thus $M_{12}$ is $5$-fold transitive and so has order $m_{12} = 12\cdot 11\cdot 10\cdot 9\cdot 8 \cdot k$, where $k$ is the order of the subgroup leaving five persons invariant. Since every permutation other than the identity of an $l$-fold transitive group must alter at least $2l-2$ symbols, it is necessary that $k$ be unity and so $m_{12} = 95040$.

As discussed in my question here, the $2l-2$ argument doesn't hold for every group (take $S_3$ as a $3$-transitive group).

But this result is only needed for a group on $12$ symbols (which is maybe what R.G. Stanton meant).

So the question arises whether any permutation of a $5$-fold transitive group on $12$ symbols moves at least $8$ symbols. This would make his argument correct.

EDIT: As discussed in the comments, $M_{12}$ is defined in terms of the automorphism group of a Steiner System $S(5,6,12)$. The "$2l-2$"- argument then only needs to apply to this group.