Here is an optional desmos graph to accompany this problem: desmos link
What I'm presenting here is my derivation of a function $H(t,T)$ which models the area-preserving continuous deformation of an arbitrary hypocycloid centered at the origin into a circle centered at the origin
Rather than thinking of $H(t,T)$ as a surface, here it should be thought of as modeling a curve which continuously changes with respect to the time $T$
I claim that the $H(t,T)$ that I have derived will work for any $n$ $\in$ $ℤ_{>2}$ that one chooses (which determines the type of hypocycloid) and any $r$ that one chooses (which determines the size of the hypocycloid)
Let $R=nr$ where $n$ $\in$ $ℤ_{>2}$
Let $f\left(t\right)=\left(\left(R-r\right)\cos\left(t\right)+r\cos\left(\frac{R-r}{r}t\right),\left(R-r\right)\sin\left(t\right)-r\sin\left(\frac{R-r}{r}t\right)\right)$
Let $A=\frac{r^{3}\sin\left(\frac{4\pi R}{r}\right)}{4\left(r-R\right)}+\pi R\left(R-2r\right)$ (This is the area enclosed by the hypocycloid, obtained using Green's theorem)
Let $a=\sqrt{\frac{A}{\pi}}$
Let $g\left(t\right)=\left(a\cos\left(t\right),a\sin\left(t\right)\right)$
Let $T$ $\in$ $[0,1]$
Let $h(t,T)=\left(1-T\right)f\left(t\right)+Tg\left(t\right)$
Let $A_{1}=\left(\left(Ta+\left(R-r\right)\left(1-T\right)\right)^{2}-\left(1-T\right)^{2}r^{2}\right)\pi$ (This is the area enclosed by the graph of $h(t,T)$, obtained using Green's theorem)
Let $H(t,T)=\sqrt{\frac{A}{A_{1}}}h\left(t\right)$
Then $$H(t,T)=\sqrt{\frac{\frac{r^{3}\sin\left(\frac{4\pi R}{r}\right)}{4\left(r-R\right)}+\pi R\left(R-2r\right)}{\left(\left(Ta+\left(R-r\right)\left(1-T\right)\right)^{2}-\left(1-T\right)^{2}r^{2}\right)\pi}}\left(\left(\left(1-T\right)\left(R-r\right)+T\sqrt{\frac{\frac{r^{3}\sin\left(\frac{4\pi R}{r}\right)}{4\left(r-R\right)}+\pi R\left(R-2r\right)}{\pi}}\right)\cos\left(t\right)+\left(1-T\right)r\cos\left(\frac{R-r}{r}t\right),\left(\left(1-T\right)\left(R-r\right)+T\sqrt{\frac{\frac{r^{3}\sin\left(\frac{4\pi R}{r}\right)}{4\left(r-R\right)}+\pi R\left(R-2r\right)}{\pi}}\right)\sin\left(t\right)-\left(1-T\right)r\sin\left(\frac{R-r}{r}t\right)\right)$$
Thus, the area-preserving continuous deformation of an hypocycloid centered at the origin into a circle centered at the origin is given by $$\boxed{H(t,T)=\sqrt{\frac{\frac{r^{3}\sin\left(4\pi n\right)}{4r\left(1-n\right)}+\pi r^{2}n\left(n-2\right)}{\left(\left(T\sqrt{\frac{A}{\pi}}+\left(1-T\right)\left(n-1\right)r\right)^{2}-\left(1-T\right)^{2}r^{2}\right)\pi}}\left(\left(\left(1-T\right)\left(n-1\right)r+T\sqrt{\frac{\frac{r^{3}\sin\left(4\pi n\right)}{4r\left(1-n\right)}+\pi r^{2}n\left(n-2\right)}{\pi}}\right)\cos\left(t\right)+\left(1-T\right)r\cos\left(\left(n-1\right)t\right),\left(\left(1-T\right)\left(n-1\right)r+T\sqrt{\frac{\frac{r^{3}\sin\left(4\pi n\right)}{4r\left(1-n\right)}+\pi r^{2}n\left(n-2\right)}{\pi}}\right)\sin\left(t\right)-\left(1-T\right)r\sin\left(\left(n-1\right)t\right)\right)}$$
where $n$ is the number of vertices that the hypocycloid has and $r$ corresponds to the size of the hypocycloid