A succinct proof that the given graphs (red $K_n$ drawn cyclically, plus blue $2$-paths between closest vertices) have dihedral automorphism groups? Take the complete graph $K_n$ ($n \geq 3$), on the red-colored vertex set $\mathbb{Z}_n$, say, and add a blue-colored $2$-path between each pair of vertices $v$, and $v+1$, we get a sequence of graphs beginning with the following:

Question: What is a succinct (of the order of one sentence) proof that the automorphism group of this vertex-colored graph is the dihedral group of order $2n$?

These graphs arise in the study of partial Latin rectangles.  If we take the partial Latin rectangle
$$
\begin{bmatrix}
1 & 2 & \cdot & \cdot & \cdot \\
\cdot & 1 & 3 & \cdot & \cdot \\
\cdot & \cdot & 1 & 4 & \cdot \\
\cdot & \cdot & \cdot & 1 & 5 \\
6 & \cdot & \cdot & \cdot & 1 \\
\end{bmatrix}
$$
and make each entry a vertex, and add an edge between vertices that share a row, column, or symbol, we get the above-right graph.  The colors arise since any autotopism or autoparatopism of the partial Latin rectangle must fix the symbol $1$.
It seems more tedious to prove the symmetries using the partial Latin rectangle directly.  So I'm hoping to shortcut this via graph theory.  But, if the graph theory proof is also tedious, there's no point.
 A: Fix $n \geq 3$ and call this graph $G$. We can distinguish between blue and red vertices, so an automorphism must send the set of blue vertices to itself. 
Label the blue vertices cyclically $v_0, \ldots, v_{n-1}$. Since $v_i$ is distance 2 from vertices $v_{i-1}$ and $v_{i+1}$ and no others (indices taken modulo $n$), an automorphism of $G$ must also preserve the cyclic ordering, or reverse it. These are precisely the dihedral group actions on $v_0, \ldots, v_n$. Since each red vertex is adjacent to exactly one pair $(v_i, v_{i+1})$, this action on the blue vertices determines the image of the red vertices as well. Therefore, any automorphism of $G$ must be given by an action of the dihedral group of order $2n$ on the blue vertices.
For the reverse direction, it's straightforward to show that any such action results in an automorphism of the graph.
Depending on how rigorous you intend to be, you might just say "imagine a blue cycle around the blue vertices, and note that the automorphisms of $G$ are exactly the automorphisms of the blue cycle."
