I have a symmetric positive definite matrix $M$ with $U$ as the Cholesky factor. There exists an upper triangular matrix $V$ where $M = V^tV$. I want to show that there exists a diagonal matrix $D$ where $diag(d)_{1:n} = 1$ or $-1$ such that $ V = DU$
Way I interpret this, there already exists a diagonal matrix $D$, now find how it fits into $A = V^tV$ and $A = U^tU$
$$ V^tV = U^tU$$ $$UV^tV = (UU^t)U$$ $$UV^tV = U$$ $$UV^t(VV^t) = UV^t$$ $$UV^t = UV^t$$
Now there exists a matrix $D$, and $V = DU$. I can show that
$$V = DU$$ $$V^t = (DU)^t$$ $$V^t = U^tD^t (D^t = D)$$ $$V^t = U^t D$$ $$UV^t = UU^tD$$ $$UV^t = D$$
Therefore the diagonal matrix $D$, is such that $V = DU$
As I understand it, you want to show that if $U$ and $V$ are two (upper triangular) Cholesky factors of $M$, then $U=DV$, where $D$ is a diagonal matrix with $\pm 1$ on the diagonal.
Indeed, let $U$ and $V$ be upper triangular matrices such that $M=V^TV=U^TU$. Then $$U^{-T}MU^{-1}=U^{-T}V^TVU^{-1}=I$$ and hence $D^TD=I$, where $D:=VU^{-1}$ is an upper triangular matrix. In order to prove the result we first show that $D$ is diagonal. This is easy since $D^TD=I$ implies $D=D^{-T}$, which says that $D$ is both upper and lower triangular and hence diagonal. Now the only (real) diagonal matrix $D$ such that $D^TD=D^2=I$ is a $D$ with $\pm 1$ on the diagonal.