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: 
Without further information about $A$, it cannot be done. The solution is not unique.

However, if $D\in\mathbb{R}^{n\times n}$ is positive semidefinite (it should be, under the circumstances), and $V\in\mathbb{R}^{p\times n}$, we can say that $A=UD^{1/2}V^T$, where $D^{1/2}$ is a square root of $D$ and $U\in\mathbb{R}^{m\times n}$ is *any* matrix satisfying $m\geq n$ and $U^TU=I$.

It looks to me that this equation came from a discussion of the singular value decomposition. Because you say that $VDV^T$ is an eigendecomposition, the diagonal elements of $D$ are actually the squares of the singular values of $A$, and the columns of $V$ are right singular vectors of $A$. The corresponding equation $AA^T=UDU^T$ would give you the left singular vectors and the same squared singular values.

A full singular value decomposition of $A\in\mathbb{R}^{m\times n}$ looks like $A=U\Sigma V^T$, where $U^TU=I$, $V^TV=I$, and $\Sigma$ is a diagonal matrix with positive entries (though not necessarily square!). In this case, then $D\_{ii}=\Sigma\_{ii}^2$, $i=1,2\dots,\min\{m,n\}$.
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.