Simplest argument that language decidable in constant time cannot be $\mathsf{NP}$-hard?
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 $P=NP$.