Undirected graph whose BFS and DFS trees have roots of degree 2
The degree of the root in a BFS tree equals the degree of the vertex in the original graph. Hence condition 2 states that your graph is 2-regular. Since it is connected, it must be a 5-cycle. Running DFS on the 5-cycle results in a path, and in particular, the root has degree 1.
This argument shows that there's nothing special about the number 5: the result remains true for any number of vertices.