Checking my understanding of Multipliers for Difference Sets In "Contemporary Design Theory: A Collection of Surveys," pg. 245 begins the section on multipliers of difference sets. 
I had previously understood a multiplier $\alpha$ of a difference set $D$ in a group $G$ to be an automorphism of $G$ which in turn induces an automorphism on the points (or blocks, if you prefer) of $\text{dev}D$, the development of $D$. (That is, $\text{dev}D$ is the symmetric design obtained by translating $D$).
I still think my understanding is correct, but the text introduces it in a weird way that makes me wonder if I am missing something. I quote the text exactly below:

"Given a difference set $D$ in $G$, it is quite often possible to obtain further automorphisms of $\text{dev}D$, not just the translations in $G$. For example, $D=\{1,2,4\}$ in $G=\mathbb{Z}_7$ describes the projective plane $PG(2,2)$ of order $2$. Clearly the automorphism $\alpha$ of $G$ given by multiplication by $2(\mod 7)$ induces an automorphism of $\text{dev}D$, since it maps the block $D+z$ to $2(D+z)=D+2z$."

Several questions arise in my mind. I hope it is okay to post them all, since they are small details related entirely to that same quote above.
1) Does an automorphism on $G$ count as a multiplier of $D$ if it induces an automorphism equivalent to a translate in $G$? Seems like this is essentially the trivial multiplier, since all the translates in $G$ have developments which are identical.
2) How does that example give an automorphism that isn't a translate of $D$? They explicitly show that its result is $D+2z$, which is clearly a translate of $D$. Similarly, why are they discussing the automorphism in the example as acting on $D+z$, when the automorphism is induced on $\text{dev}D$? 
3) Shouldn't a non-trivial multiplier yield a new difference set $D'$ whose deveolpment is a distinct symmetric design from $\text{dev}D$?
I can tell that my confusion hinges on something small and crucial. Thanks very much for any assistance you can provide.
 A: After several discussions with a few colleagues, I have been able to answer my own question. I will do so, then accept it, so that it does not add to the unanswered queue. 
If anyone stumbles upon this and has issues with the subject, please ping me. I'd be happy to bounce ideas off of another person. I imagine this would benefit both of us.
My original understanding of a multiplier is correct. The key (the small bit I knew I was missing) is that not all automorphisms of $G$ induce automorphisms on $\text{dev}D$, and vice-versa.
Additionally, the phrase "automorphism of $\text{dev}D$" is really an abuse of language. Really, the automorphism is on the point or block set (these are equivalent as the linked text shows earlier), though it is frequently more convenient to think of the block set of the design as being permuted. Instead of saying "automorphism of $\mathcal{B}$, the block set of $\mathcal{D}=\text{dev}D$, we just abuse the language and say "automorphism of $\mathcal{D}$".
1) Yes, it does, and in fact this is the translate by the zero element. It is also the only translate that counts as a multiplier, because it is the only one preserving the group operation. More on this below.
2) I was asking the wrong sort of thing here. The translates of $D$ are automorphisms of the block set of $\text{dev}D$. They are not automorphisms of $G$, which all multipliers must be, because they do not fix the identity element. It is easy to verify this when the group operation is addition, for example. As for the rest of the question, the answer seems to be that math textbooks do not always give the best presentations.
3) In the words of my advisor: "No! Absolutely not!" Although I admit this question arose from me copying something down wrong and playing with it, leading to nonsensical results.

Here is a rehashing of the example in the book which may help others. I added a lot of side detail that I found helpful: $D=\{1,2,4\}$ in $G=\mathbb{Z}_7$ describes the projective plane $PG(2,2)$ of order $2$. That is, by translating $D$, we obtain the $(7,3,1)$-design known as the Fano plane. 
Those translates are automorphisms of $\text{dev}D=PG(2,2)$ because they preserve the block structure. It is easy to check that $\text{dev}(D+1)=PG(2,2)$ as well, for example. But these are not automorphisms of $G=\mathbb{Z}_7$ because they do not preserve the group operation. Again, this is easy to check.
However, the automorphism defined by $\sigma_a(x)=ax$, where $x\in G$, is in fact sometimes an automorphism of $\text{dev}D$ as well. Writing it out shows that if the blocks of $\text{dev}D$ are written as translates such that block $i$ is $D+i$, then $\sigma_2=(253)(467)$, where the permutations are on the indices of the blocks.
These are called multipliers because, for example, $\sigma_2$ is just literal multiplication by $2$. We say $2$ is a multiplier of $D$ in this case. By the way, it is easy to check that $3$, for example, is not a multiplier of $D$, because $\sigma_3(D)=\{2,5,6\}$, which is not a block in $\text{dev}D$, so $\sigma_3$ does not preserve blocks, and is therefore not an automorphism on $\text{dev}D$ (despite being one on $G$). Mutlipliers must be automorphisms on both.
What a hoot!
