Simplest argument that language decidable in constant time cannot be $\mathsf{NP}$-hard? My question is specifically about $\emptyset$, but more generally about any language that can be decided in (deterministic or nondeterministic doesn't really make a difference here) constant time. Obviously, $\emptyset$ cannot be $\mathsf{NP}$-hard since there is nothing to map "yes"-instances to, but I'd prefer a one line argument using a time hierarchy theorem, ideally. Currently I'm arguing (very informally speaking) that since constant time is $\subsetneq \mathbb{o}(n \log n)$, constant time is strictly "less powerful" than any deterministic polynomial time due to the deterministic time hierarchy theorem and thus - since $$\mathsf{Polytime}(f(n)) \subseteq \mathsf{NPolytime}(f(n))$$ - also strictly less powerful than nondeterministic polynomial time. So, constant time languages can't be $\mathsf{NP}$-hard.

This doesn't sound elegant to me at all (although it should be correct I hope, even if not phrased as formally as I'd write it down for something to hand in). What's the most elegant argument you can come up with, preferably using not-too-advanced complexity theory?
Let $L \in P$ be non-trivial (not empty and not containing everything). If $L$ is NP-hard then $P=NP$. Indeed, given any language $L' \in NP$, by definition of NP-hardness there is a polytime reduction $f$ such that $x \in L'$ iff $f(x) \in L$. Since $L \in P$, this gives a polytime algorithm for $L'$, showing that $L' \in P$.

Conversely, if $P=NP$ then $L$ is NP-hard. Indeed, given $L' \in NP$, we can decide $L'$ in polynomial time using some algorithm $A$. Fix some yes instance and some no instance of $L$, and use $A$ to come up with a polytime reduction from $L'$ to $L$, which always returns one of the two fixed instances. This shows that $L$ is NP-hard.

Summarizing, if $L \in P$ is non-trivial then $L$ is NP-hard iff $P=NP$.