I conjecture the following :
If $G$ is a graph so that every pair of adjacent vertices have different degree, then $G$ is not asymmetric graph. I will remind that a graph is a asymmetric if there are no automorphisms, except for the trivial automorphisms.
It is possible that this statement is incorrect. Can someone find a counterexample?
And a similar conjecture:
If $G$ is a $2$-connected chordal graph, then it contains a triangle where at least two vertices are of the same degree.
$K_1$ is an asymmetric graph in which adjacent vertices have different degrees. Here is a less trivial example.
Start with an asymmetric tree $T$ with degree sequence $3,2,2,2,1,1,1$.
The complement of $T$ is an asymmetric graph $\overline T$ with degree sequence $5,5,5,4,4,4,3$.
Subdividing each edge of $\overline T$, we get an asymmetric bipartite graph $G$ with degree sequence $5,5,5,4,4,4,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2$ in which each edge joins a vertex of degree $2$ to a vertex of degree $\ge3$.
Here's a smaller example: take a path with vertices $v_0,v_1,v_2,v_3,v_4,v_5,v_6$ and edges $v_0v_1,v_1v_2,v_2v_3,v_3v_4,v_4v_5,v_5v_6$, and add two more edges $v_1v_3$ and $v_1v_5$.