If $A(x)^{T}A(x) = C$, then $A(x) = O\widetilde{C}$ with $O$ Orthogonal?

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Let $A(x)$ be an $n \times m$ matrix-valued function of the vector variable $x \in \mathbb{R}^n$, and let $C \in \mathbb{R}^{m \times m}$ be a fixed positive semidefinite matrix. Is the following bidirectional statement true?

$$ A(x)^{T}A(x) = C\;\;\;\forall x \qquad \Longleftrightarrow \qquad A(x) = O(x)\tilde{C} $$ for some $n \times n$ orthogonal matrix $O(x)$ some $\tilde{C} \in \mathbb{R}^{n \times m}$ such that $\tilde{C}^{T}\tilde{C} = C$.

(The $\Longleftarrow$ implication is obvious; my question really concerns the $\Longrightarrow$ implication).

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Let $A=VDW$ be the singular value decomposition. If you re-write the equality $A=VDW$ as $A=(VW)W^TDW$, we get the polar decomposition $A=UM$, with $M$ positive-semidefinite and $U$ orthogonal. Since $$ M=(M^TM)^{1/2}=(A^TU^TUA)^{1/2}=(A^TA)^{1/2}=C^{1/2}, $$ the matrix $M$ does not depend on $x$.