I'm trying to wrap my head around the idea of automorphisms, and I'm having a lot of issues.
One of the questions I've been given as an exercise is thus;
Let $\mathbb{V} = \{1, 2, 3, 4, 5, 6\}$ and $\mathbb{B} = \{ \{1, 2, 3\}, \{1, 5, 6\}, \{3, 4, 5\}, \{2, 4, 6\} \}$. Determine $|Aut(\mathbb{V}, \mathbb{B})|$.
With this, I'm a little confused. I've noted that, at least initially, the permutation shifting the blocks is $(1,3,5)(2,4,6)$. If I fix the element $1$, then the permutations $(2,5)(3,6)$ and $(2,6)(3,5)$ also map the blocks to themselves. If I try to fix two elements, I find that I can't find any such automorphism. But, I'm not sure where to go from here.
I'm under the impression that I have to use the Orbit-Stabilizer Theorem, but again, my implementation of this is a bit dodgy. If anyone could point me towards some information or areas in which I could read about this topic of group theory within the context of designs, that'd be much appreciated.
You could show that there are only four automorphisms fixing $1$, namely, $(2,5)(4,6)$, $(2,6)(3,5)$, $(2,3)(5,6)$ and the identity. These together with $(1,3,5)(2,4,6)$ generate a transitive group (i.e. there is a single orbit), so by the Orbit-Stabilizer Theorem, the full automorphism group $G$ has order $6 \times 4 = 24$.
It is not clear what exactly they mean by "determine". The three automorphisms you have found generate $G$, so you could say that you have determined it. It would be tedious if they expected you to write down all $24$ permutations.