Consider two projections $p,q\in M_n(\mathbb{C})$ (i.e. $p^2=p=p^*$ and the same for $q$) with $\operatorname{rank}(p)=\operatorname{rank}(q)$.
Claim: There exists a $v\in M_n(\mathbb{C})$ such that $p=v^*v$ and $q=vv^*$.
I don't know how to define such a $v$. What I have tried: Since $p$ and $q$ are self-adjoint, there exist unitaries $x,y\in M_n(\mathbb{C})$ such that $x^*px=d_1$ and $y^*qy=d_2$, where $d_1$ and $d_2$ are diagonal matrices with eigenvalues $\in \{0,1\}$. Furthermore, $\operatorname{rank}(p)=\operatorname{rank}(q)$ implies $d_1=d_2$. One of my attemps was to define $v=yx^*p$, but this doesn't work. With this definition we have $vv^*=yx^*ppxy^*=yx^*pxy^*=yd_1y^*=yd_2y^*=q$, however, we don't have $v^*v=p$.
Thus my question is: How to define $v$ correctly?
Let's alter your definition a little bit, let $v := qyx^*p$, then $$ vv^* = qyx^*pxy^*q = qyd_1y^*q = qyd_2y^*q = q^3 = q $$ and $$ v^*v = pxy^*qyx^*p = pxd_2x^*p = pxd_1x^*p = p^3 = p. $$