I am trying to answer this question and any method I can think of requires a knowledge of $A^TA$ given that $AA^T=D$, where $D$ is diagonal and $A$ is a square matrix. I could not find anything useful in MSE or elsewhere and I was unable to do any progress by myself.
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All you can say about $A^{\mathsf T}A$ is that is a symmetric matrix similar to $AA^{\mathsf T} = D$ (that is, it diagonalizes to $D$).
We can see this from $A$'s singular value decomposition: if $A = U\Sigma V^{\mathsf T}$, then $AA^{\mathsf T} = U \Sigma^2 U^{\mathsf T}$ and $A^{\mathsf T}A = V\Sigma^2 V^{\mathsf T}$, so both are similar to $\Sigma^2$.
To see that we cannot do better, note that if we replace $A$ by $AQ$, where $Q$ is any orthogonal matrix, then $(AQ)(AQ)^{\mathsf T} = AQQ^{\mathsf T}A^{\mathsf T} = AA^{\mathsf T} = D$, while $(AQ)^{\mathsf T}AQ = Q^{\mathsf T}(A^{\mathsf T}A)Q$, which is the conjugation of $A^{\mathsf T}A$ by an arbitrary orthogonal matrix $Q$.
In the case that $A$ is not necessarily square, we cannot even say this much, although from the singular value decomposition we can still conclude that $AA^{\mathsf T}$ and $A^{\mathsf T}A$ have the same nonzero eigenvalues.
Misha Lavrov
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1Although your answer is great, the @Strants' answer is more useful for my purposes, so I had to accept his answer even after I had accepted yours. Thank you. – Joca Ramiro Mar 06 '18 at 17:19
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No problem: that's what accepting answers is for. – Misha Lavrov Mar 06 '18 at 18:08
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If you are expecting $A^\top A$ to be diagonal, here is a counterexample. Let $A = \begin{bmatrix}1 & -1 \\ 2 & 2\end{bmatrix}$. Then $AA^\top$ is diagonal but $A^\top A$ is not.
angryavian
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@tobwin $\left[\begin{smallmatrix}1&-1\2&2\end{smallmatrix}\right]\cdot\left[\begin{smallmatrix}1&2\-1&2\end{smallmatrix}\right]=\left[\begin{smallmatrix}2&0\0&8\end{smallmatrix}\right]$. Multiplying in the other order though gives $\left[\begin{smallmatrix}5&3\3&5\end{smallmatrix}\right]$ – JMoravitz Mar 06 '18 at 16:59
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This example actually works over any commutative ring, although $A=\pmatrix{1&1\ 0&0}$ would make a simpler example. – user1551 Mar 06 '18 at 17:59
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If $A^TA = D$, then the columns of $A$ are orthogonal. Hence, $A = O\Lambda$ for some diagonal matrix $\Lambda$. (In particular, if $A$ is nonsingular, we can take $\Lambda_{ii}$ to be the norm of the $i$th column of $A$, and let the $i$th column of $O$ be a normalized version of the $i$th column of $A$.)
Then, $$A^TA = \Lambda^TO^TO\Lambda = \Lambda^2 = D$$ so $\Lambda$ is the positive square root of $D$. Thus, $$AA^T = O \Lambda \Lambda^T O^T = O D O^T.$$