How to determine all graph automorphisms for a given graph? In Farb and Margalit's A Primer on Mapping Class Groups, the proof of Theorem 3.10 uses the notion of graph automorphisms. I don't know much about graph theory and have trouble following their argument.
Consider the following graph $\Gamma$:


[...], the only automorphisms of $\Gamma$ come from flipping the three edges that form loops and swapping pairs of edges that form a loop. In particular, any automorphism of $\Gamma$ must fix the three edges coming from $\alpha_4$. Thus, we see that $\phi$ preserves the orientation of $\alpha_4$, [...]

The edges of this graph come from simple closed curves on a surface $S$ and the vertices are the intersections. $\alpha_4$ is the curve represented by the loop containing three edges and $\phi$ is an orientation-preserving homeomorphism on $S$ fixing $\Gamma$.
Questions 1: What is meant by flipping the three edges? Permuting them? Rotating them? Something else entirely?
Question 2: How can I convince myself that those automorphisms are the only automorphisms?
Question 3: Usually in the book, if a map $f$ fixes a set $M$, usually $f(M) \subseteq M$ is meant. Do Farb and Margalit mean that each edge of $\alpha_4$ is fixed individually here? That would make more sense in the context, but not coincide with the usual use of this notion in the book.
 A: Q1. The text you're quoting from seems to be confused -- from a graph-theoretic point of view "flipping" one of the loop edge is not a meaningful operation. Intuitively it sounds like it would mean reversing the direction of the edge, but that's not a change that's visible at the graph level.
Since the graph is a multigraph, presumably an automorphism will be a map that takes vertices to vertices and edges to edges -- but "flipping' an edge would still map the edge to itself, and the automorphism would simply be the identity map.
It is possible that the book uses a strange variant notion of "graph morphism" -- but then it really ought to provide a precise definition that you can check.
Q2. Each of the three vertices with a self-loop must map to some vertex with a self-loop. But exactly one of those vertices has two distinct neighbors, so it must map to itself. The other two self-loop vertices have different distances to the one we just fixed, so each of them must also map to itself. Each of the remaining vertices have a unique combination of distances to the three ones we have fixed so far, so must also map to themselves.
So any automorphism must fix all of the vertices, and the only choice there is in making one is, for each pair of parallel edges, whether it swaps those edges or not.
Q3. I'm not sure what $\alpha_4$ actually refers to, but yes: to "fix a set" is an ambiguous phrasing, and you may have to depend on context to figure out which of the possible meanings make sense in each case. (I wouldn't say that $f(M)\subseteq M$ is a common meaning of "fixing $M$", but $f(M)=M$ certainly is; as is $\forall x\in M: f(x)=x$).
