homeomorphic two spheres embedded in $\mathbb{R}^4$

Question

Let $A$ and $A'$ be two annuli in $\mathbb{R}^3$. Suppose $A$ has $n$ half twists and $A'$ is with $m$ half twists, where $m$ and $n$ are even and $m\ne n$. It is clear that the surface resulting by cutting along the two boundary circles of an annulus is a 2-sphere. Let $S$ and $S'$ denote the two spheres resulting from the surgery explained above done to $A$ and $A'$, respectively. A 2-knot is a sphere embedded in the Euclidean 4-sapce $\mathbb{R}^4$. By embedding $S$ and $S'$ in $\mathbb{R}^4$ we obtain 2-knots denoted by $F$ and $F'$ respectively. We consider the projection of the 2-knot into $\mathbb{R}^3$ which gives a generic surface with singularity set consists of double points, triple points or branch points. My question is: When we can say $F$ and $F'$ are equivalent in $\mathbb{R}^4$, in other word when their projections in $\mathbb{R}^3$ are isotopy? I know the Roseman moves gives the answer but here I am focusing on the twist part, I don't know whether it has an affect or not.