That's a good question, but you're not correct.
The reason is that if $f\colon\varnothing\to S$, then $f=\varnothing$. Therefore $\breve f=\varnothing$. Therefore $f=\breve f$ and both satisfy the condition for being a function.
It is true, however, that $\breve f$ is not a function whose domain is $S$ (unless $S$ is empty). And note that we shouldn't require it is a function from $S$, because then the function $f(n)=n+1$ as a function from $\Bbb N$ to itself is not injective anymore, which is plain preposterous.
Also, there is but one unique empty function. The empty set.