Yes, there is.
(As said in the comments) the function $f=\emptyset$ is the bijection you seek.
It may seem strange that $f$ is a set, but this is ok since the definition of a function is a set of pairs (s.t ...), in this case we take $f$ to be an empty set.
The definition of $f$ is fine since there are no $x, y_1,y_2 \in \emptyset$ s.t $y_1\neq y_2$ and s.t $(x,y_1),(x,y_2)\in f$.
$f$ is injective since there are no two different elements in $\emptyset$ that $f$ maps to the the same element (this is just because there are no two different elements in $\emptyset$).
$f$ is surjective since for each $y$ in $\emptyset$ there exist a source (this is since there are no elements in $\emptyset$ hence we can't say there exist an element that doesn't have a source).