automorphism in Cai-Furer-Immerman gadgets
You're missing the emptyset $\emptyset$ which is connected to all $b$'s. To get an  automorphism, you select a subset $T\subseteq \{1,...,k\}$ of even cardinality and then swaps $a_i$ with $b_i$ for each $i\in T$ and then adjusts the sets in the middle. In your example the graph is $$(a_1,\{12\}),(a_2,\{12\}),(b_1,\emptyset),(b_2,\emptyset).$$

Still in your example if $T=\emptyset$ you don't need to do anything and if 
$T=\{1,2\}$ the automorphism is given by swapping $a_1$ with $b_1$, $a_2$ with $b_2$ and $\{1,2\}$ with $\emptyset$.

Now for the general case, we need to show that there is always a way of adjusting the middle vertices. We know that $T$ has even cardinality. So let $|T|=2r$. We just need to show that such an 
automorphism exists if $|T|=2$ since otherwise we can apply the composition of $r$ automorphisms corresponding to partitioning $T$ into $r$ subsets of size $2$. Thus assume $T=\{i,j\}$.  Then the automorphism swaps $a_i$ with $b_i$, 
$a_j$ with $b_j$, each middle vertex $S$ such that $S\cap\{i,j\}=\emptyset$ with the middle vertex $S\cup \{i,j\}$ (this can be seen in your example), and each subset $S$ such that $S\cap \{i,j\}=\{i\}$ with the subset such that $S\cap \{i,j\}= \{j\}$ (This you can see for $k=3$). Notice that this swapping process is an automorphism since for an index $p\neq \{i,j\}$ the edge relation between $a_p$, $b_p$ and these swapped vertices is completely preserved, and clearly the edge relation between $a_i,a_j,b_i,b_j$ is properly adjusted. 

Finally to see that these are the only possible automorphisms, notice that each $a_i,b_i$ is colored with its own color. So they cannot be mapped to another pair $a_j,b_j$. Also notice that it is not possible to have an automorphism that maps an middle vertex to a middle vertex without swapping some $a_i$ with some $b_j$. $\square$