Question: I am familiar with solving matrix equations by multiplying by the inverse of a matrix (if it exists), but how can I solve a matrix equation for $A$, if it contains $A^TA$?

As a specific example:

Solve for A: $$A^TA = VDV^T$$

(Note: $VDV^T$ is an eigendecomposition, so $D$ is diagonal. I'm not sure if that's useful.)

Answer: 
In general this will not be unique because even under the friendly condition that $A$ is symmetric, then we could get non-unique answers for $A$. For example, 

$$
\begin{bmatrix}
 -1 & 0 \\
 0 &-1
\end{bmatrix}
\begin{bmatrix}
 -1 & 0 \\
 0 &-1
\end{bmatrix}
=
\begin{bmatrix}
 1 & 0 \\
 0 &1
\end{bmatrix}
\begin{bmatrix}
 1 & 0 \\
 0 &1
\end{bmatrix}
= 
\begin{bmatrix}
 1 & 0 \\
 0 &-1
\end{bmatrix}
\begin{bmatrix}
 1 & 0 \\
 0 &-1
\end{bmatrix}
=
\begin{bmatrix}
 -1 & 0 \\
 0 &1
\end{bmatrix}
\begin{bmatrix}
 -1 & 0 \\
 0 &1
\end{bmatrix}
$$

So they all solve 
$$A^TA=VDV^T$$
In the specific case you mention, let $D\_0$ be some diagonal matrix such that $D\_0^2 = D$. Then A = $D\_0 V^T$ is a solution. I suppose that this suggests some sort of general approach: get $A^T A$ on one side of the equation, and then try to break the other side into a matrix and its transpose. I can imagine this being hard in general, though in your particular question it is very easy.

Another approach would be to use the fact that you can decompose a symmetric matrix as $U^T D U$ where U is unitary and D is diagonal and note that $A^TA$ is symmetric. This would again, not likely resolve all questions of this nature.