If O(log n) vs O(n) is exponential what is O(1) vs O(n)?
There's no name for that, but the closest would be "infinite" speedup. $O(\log n)$ is considered an exponential speedup over $O(n)$, because $n$ is exponential in $\log n$: if we take the exponential function $f(n)=2^n$, we get $f(\log n)=2^{\log n}=n$.

However, $n$ isn't anything in $1$: there's no way to recover $n$ from $1$, no matter how quickly the function $f$ that we take grows, $f(1)$ will always be a constant (and never equal $n$). As $n\to \infty$, $1$ becomes infinitely smaller than $n$.