Let $V \subseteq \mathbb{CP}^n$ be a projective variety embedded in complex projective n-space. By the nullstellensatz it is the zero-set of a finite number of homogeneous polynomials, $p_1(x_1,...x_{n+1}),...,p_k(x_1,...x_{n+1})$, in variables $x_1,...x_{n+1}$, which are coordinates in $\mathbb{CP}^n$. Let $PGL(n)$ be the group of projective transformations of $\mathbb{CP}^n$, whose action on $\mathbb{CP}^n$ is induced, as usual, by the action of $GL(n+1)$ on $\mathbb{C}^{n+1}$. We say that two such varieties $V,U \subseteq \mathbb{CP}^n$ are projectively equivalent if one can be transformed to the other by a projective transformation (i.e. they are in the same orbit of the action of $PGL(n)$ on subsets of $\mathbb{CP}^n$).
As explained for example here and here, given a finite set of homogeneous polynomials $p_1(x_1,...x_{n+1}),...,p_k(x_1,...x_{n+1})$, the algebra of invariants of these polynomials is the (graded) algebra of all those polynomials in the coefficients of the original polynomials, which are invariant, up to a multiplicative factor, under the action of $GL(n+1)$ on $\mathbb{C}^{n+1}$. In addition, that constant factor has to be a power of the determinant of the transformation.
In other words, the invariants are polynomial functions in the coefficients of the original polynomials, such that when one substitutes in the original polynomials a linear change of variables, the invariants stay the same, up to a factor which is a power of the determinant of the linear change of variables. Similarly (as explained in the references), one also has the algebra of covariants of a set of polynomials, which is defined similarly except that the covariants are polynomial functions in the coefficients of the original polynomials and in the variables themselves. According to Hilbert's Basis Theorem, both algebras are finitely generated.
Now, invariants of a set of homogeneous polynomials are not enough to characterize the corresponding embedded projective variety up to projective equivalence. For example (as described in both references), for a binary cubic, i.e. a homogeneous polynomial of degree 3 in 2 variables, the algebra of invariants is generated by the discriminant, but the discriminant is 0 if the cubic has a double root and a single root, or if the cubic has a triple root. The first case corresponds to the cubic representing a variety in $\mathbb{CP}^{1}$ consisting of two distinct points, while the second case corresponds to the cubic representing a variety in $\mathbb{CP}^{1}$ consisting of a single point. These two are clearly not projectively equivalent.
However, the above book by Olver also mentions that the invariants and covariants together are enough to characterize the corresponding sets of points up to projective equivalence. Specifically, a binary cubic has its hessian as a covariant; while one cannot speak of the value of the hessian of a binary cubic (since it is a function of $x_1,x_2$ as well as being a function of the coefficients), one can speak of the hessian being identically zero or not - and more generally, of two hessians being equal, up to a factor (by which I mean that one hessian is a multiple of the other by a number). Olver shows that the hessian vanishes identically iff the cubic has a triple root; thus the discriminant and hessian together completely characterize the set of zeros of a binary cubic up to projective equivalence - if the roots are distinct then both are nonzero; if there is a double root then the discriminant vanishes but the hessian does not; and if there is a triple root then both vanish.
My question, roughly, is whether this idea generalizes to arbitrary embedded varieties. Are the algebras of invariants and covariants of the defining polynomials, together, enough to characterize the embedded variety up to projective equivalence?
To state the question precisely, we need to first introduce absolute invariants - these are invariants that do not change at all under a projective transformation of the coordinates (i.e. even a multiplication by a power of the determinant is now not allowed), but they are allowed to be rational functions of the coefficients of the original polynomials - not only polynomials in them. The first kind of invariants that we discussed are now called relative invariants. It is easy to realize and see that the field of absolute invariants of a finite set of polynomials is generated by certain quotients of powers of the generators of their algebra of relative invariants (basically you take quotients of two generators, each raised to a power, such that the resulting quotient is unchanged by a projective transformation because the determinantal powers in the numerator and denominator cancel each other). Similarly one has absolute covariants and their field, with its generators.
Now, say we are given two finite sets of homogeneous polynomials over $\mathbb{C}$, $p_1(x_1,...x_{n+1}),...,p_k(x_1,...x_{n+1})$ and $q_1(x_1,...x_{n+1}),...,q_k(x_1,...x_{n+1})$, both having the same number of polynomials, and with the polynomials having the same degrees $d_1,...d_k$. Take a generating set $b_1,...b_m$ for the field of absolute invariants of a set of $k$ polynomials of degrees $d_1,...d_k$, and take a generating set $c_1,...c_l$ for the field of absolute covariants of a set of $k$ polynomials of degrees $d_1,...d_k$. Suppose that every invariant in $b_1,...b_m$ has the same value for the polynomials in $p_1(x_1,...x_{n+1}),...,p_k(x_1,...x_{n+1})$ and for those in $q_1(x_1,...x_{n+1}),...,q_k(x_1,...x_{n+1})$ (that is, when the coefficients of these polynomials are substituted in the rational functions $b_1,...b_m$, the same complex numbers are obtained). Suppose also that every covariant in $c_1,...c_l$ is equal for the polynomials in $p_1(x_1,...x_{n+1}),...,p_k(x_1,...x_{n+1})$ and for the polynomials in $q_1(x_1,...x_{n+1}),...,q_k(x_1,...x_{n+1})$ (equality here is not equality of complex numbers, but equality in a certain field of rational functions). Does it follow that the two embedded varieties defined by $p_1(x_1,...x_{n+1}),...,p_k(x_1,...x_{n+1})$ and $q_1(x_1,...x_{n+1}),...,q_k(x_1,...x_{n+1})$ are projectively equivalent?
(technically one should also consider the case when some of the absolute invariants and covariants are undefined since the substitution causes some denominators of the rational functions which define them to be zero, but this is already complicated enough so let's just forget about this case for now.)
Another related question that naturally follows from this is whether the invariants only are enough, if we restrict ourselves to irreducible varieties. In the above exmaple of a binary cubic, the covariant (the hessian) is only needed to distinguish between the two degenerate cases that result from the defining equation being reducible (a double root and a triple root). If we restrict ourselves to varieties defined by irreducible polynomials, are the values of the invariants in a generating set $b_1,...b_m$ for the field of absolute invariants enough to determine the variety up to projective equivalence? Further related questions are what happens when we look at varieties/polynomials above $\mathbb{R}$ instead of $\mathbb{C}$? Above other fields? And what happens when we look at equivalence relative to subgroups of $GL(n+1)$ instead of the full group (such as affine equivalence or euclidean equivalence) - since as is well known, such subgroups also have their corresponding algebras of invariants? I welcome any partial answer or direction to answering any of these questions.
Resources on classical invariant theory are notoriously difficult to find, and I was somewhat surprised that this question is not addressed in any resource that I know (even though Olver's book above comes very close to discussing this question, but then backs away). It seems like a question that would have been pretty interesting to the projective geometers of the 19th century. For this reason, if anyone knows a nice reference that discusses the relation of classical invariant theory to projective geometry, I'd also like to know (Salmon's classic treatises don't seem to answer this question).