Number of symmetries of some Diagrams appearing in QFT I would like to get the number of symmetries (or the order of the automorphism group) associated to some (vacuum) graphs arising in scalar $\phi^4$-theory .

My efforts so far are:
First graph:
we have a $Z_2$ symmetry associated with exchanging the two bubbles. $Z_2$ has order two which gives us a factor of $2$. Furthermore  we have two addiontal $Z_2$ symmetries which corresponds to a flipping of bubbles around an axis passing through the vertex perpendicular to the bubbles. This gives us a $Z_2 \times Z_2$ symmetry which has order $2\cdot2=4$ and therefore the complete diagramm has $Z_2 \times Z_2 \times Z_2 $ symmetry and therefore $\text{Aut}(g)=8$. Is this reasoning correctt because it seems to fail in the next example (the number in the picture are the correct ones).
Second graph:
We can exchange the three bubbles as we want which results in a symmetry group $S_3$ which has order $3!$. Furhtermore we might flip any of the bubbles around the same axis as described above yielding an additional  $Z_2 \times Z_2 \times Z_2 $  symmetry which has oder $8$ and therefore we have $\text{Aut}(g)=48$ which is of by a factor of three. Where have i go wrong here (the 3 suggests somehow that it should be $S_2$ and not $S_3$)?
Third graph:
We get two $Z_2 \times Z_2 \times Z_2$ symmetries from the single objects and an additional $Z_2$ for exchanging the two. Makes  $\text{Aut}(g)=8\cdot 8\cdot 2=128$. Here everything seems to work out fine so far
Fourth graph:
We might interchange any of the four lines which is a $S_4$ symmetry which gives us a $4!$. Furthermore we have an $Z_2$ symmetry corresponding to interchange of the two verticies. Therefore $\text{Aut}(g)=24\cdot 2=48$. Nice!
What i would kindly ask you is the following:

1.) Where goes my reasoning wrong in the calulations of the second diagram
2.)  Is the reasoning for the other diagrams correct?

My background:
Group theory: On the level of an average theoretical physicist. Lots of heuristics but not a lot of formal education.
Graph theory:
Almost nothing, besides the stuff one learns in QFT (this is not much)
 A: Let's just start with the definition of a graph automorphism. Given a simple (no multiple edges, no `loops') graph with vertex set $V$, an automorphism is just a function $f \colon V \to V$ that preserves incidence. So if $v_1, v_2 \in V$ are any pair of vertices, we must have $f(v_1)$ and $f(v_2)$ connected by an edge if and only if $v_1$ and $v_2$ are connected by an edge.
It's generally not easy to find graph automorphisms, but fairly easy to understand what they are, just ways to shuffle the labels of the graph. But it gets trickier to define automorphisms for graph with multiple edges. Let's define the underlying simple graph for any non-simple graph as the simple graph obtained identifying all edges with the same endpoints, and collapse any loops down to the vertex they come from.
For the graphs you've given, their underlying simple graphs and automorphism groups (of the underlying simple graph) are listed in the following table.
\begin{array}{|r|c|c|c|c|}
\hline\text{example:} & 1 & 2 & 3 & 4 \\
\hline \text{underlying simple graph}: & \bullet & \bullet - \bullet - \bullet & \bullet - \bullet\ \ \bullet - \bullet & \bullet - \bullet \\ \hline
\text{automorphism group:} & Z_1 & Z_2 & Z_2^3 & Z_2 \\
\hline
\end{array}
The reason $\bullet - \bullet - \bullet$ has only $Z_2$ as a symmetry group and not $S_3$ is because the middle vertex can't get moved anywhere: Automorphisms preserve the degree (the number of edges connected) of a vertex, and it's the unique vertex of degree $2$. So we can only swap the outer vertices, or do nothing.
So in order to calculate the number of automorphisms of your graphs, we do two things:


*

*Find the number of automorphisms of the underlying simple graph.

*Keep track of all "bundles" of edges. A "bundle" of edges is a set of edges that all have the same endpoints (we identify all edges in a bundle into a single edge, in the underlying simple graph). We can have bundles of loops too. Each bundle containing $k$ edges contributes a factor of $S_k$ to the symmetry group of the overall graph, as we can permute these edges however we like.
The whole automorphism group contains the product of the above groups. Normally, that would be the whole automorphism group, but ...
The only odd thing (mathematically) is that for you, it appears that each loop contributes a factor of $Z_2$ to the symmetry group (I'm sure this happens for a perfectly good QFT reason, like traversing the loop backwards or something). I would normally say the first graph has an automorphism group of $Z_2$ from permuting the two loops that form a bundle, but it is evidently
$$\underbrace{Z_2}_{\text{permuting loops}} \times \underbrace{Z_2 \times Z_2}_{Z_2 \text{ from each loop}}$$
Your arguments seem to take all of this into account perfectly, except you'd misidentified the automorphism group of $\bullet - \bullet - \bullet$ as $S_3$, rather than the $Z_2$ that it is.
