Since any group of order 3⋅2n3⋅2n has ∣G∣≥3∣G∣≥3, it cannot admit a Cayley graph which is a tree. Hence:

    No Cayley graph of a group of order 3⋅2n3⋅2n can have a unique path between every pair of vertices.

My mistake, I said unique path when I should have said unique shortest path.

Also, there are trivial solutions with odd cycles and complete graphs which must be excluded. (So the answer to the prompt as originally stated is wrong too)