$4$-regular graph with exactly one perfect matching Can there be a $4$-regular graph with exactly one perfect matching? That is a graph that does have a perfect matching, but not two (not necessarily disjoint) perfect matchings.
Based on a well-know result due to Kotzig, a graph with a unique perfect matching has a cut edge (see for example the book: Matching Theory by Lovasz and Plummer). But a 4-regular graph cannot have a cut edge, so it cannot have a unique perfect matching.
EDIT 2020-04-13: this suggestion of how to give an easier proof is wrong.  I am sorry. I will leave it up as it might be interesting nonetheless. I'll strike the wrong statement. (Of course, obviously, 3-regular graphs can contain a bridge. My apologies.)

Ghodrati has already given a correct proof. It is worth pointing out 


  it is also possible to deduce this from Petersen's classic theorem on that every bridgeless cubic graph has a perfect matching. 


(Arguably, this is a more lightweight tool that Kotzig's theorem used by Ghodrati; that's the main reason why I make this remark). 

To see this, simply note that if there were a $4$-regular graph with a unique perfect matching, then removing this matching would leave a $3$-regular graph without a perfect matching; by Petersen's theorem, it would have to have a bridge;  but a $3$-regular graph obviously cannot have any bridge, as is easily seen by summing degrees. This proves that a $4$-regular graph with a unique perfect matching is impossible.