As it turns out though, your problem is actually quite simple.
Assume that V is full rank. As long as that is true, then as can arbitrarily make the substitution
X = A*V
V is known and of full rank, so if we can solve the simpler quadratic form
X'*B*X = C
then we can trivially recover A from X.
I'll note there is much to be found in the literature, solving what seems to be a similar equation, X*A*X=B. For example:
Of course, your problem is different because of the transpose. And that is what makes the solution trivial. Lets take this a step further. B is symmetric and square. Is it positive definite? If so, then we can write B as
B = L'*L
So a Cholesky decomposition of B. Now make a further transformation, with
Y = L*X
Your problem now becomes to solve for Y in the problem
Y'*Y = C
Again, is C positive definite? If so, then we can write Y in the form of a Cholesky decomposition of C. Once you have Y, recover X. Once you have X, recover A.
So the problem becomes trivial, if B and C are positive definite, and V is any full rank matrix.
Edit: Since I see by your response to Matt that V is non-square, the problem is still relatively easy, although the solution now becomes far less unique. If
X = A*V
then once X is known, recover a solution for A as
A = X*pinv(V)