Confusions about Covariant and Contravariant vectors I am trying to connect the concepts I learned from special relativity, to those of general relativity.  Take a look at this example from wikipedia.  They find a transformation matrix from the contravariant components of a vector, to the covariant components.
Now let's move to general relativity.  I know that in flat space, the metric tensor is just the Minkowski metric $\eta_{\mu\nu}$, and I know that in order to change a vector to a covector, you simply contract the metric with the vector. 
But if I were to take a vector $V^{\mu}$ and lower the index to a covector $V_{\mu}$ in flat space, it most certainly would not be the complicated change of basis matrix shown in the example.  Am I missing something here? When  you lower an index, are you finding an entirely different entity? Or are you finding the covariant components of the same vector? 
I hope this makes sense.
 A: 
When you lower an index, are you finding an entirely different entity?

Yes, it's a different entity. $V^\mu$ are the components of the vector $\vec{V}$ while $V_\mu$ are the components of the one-form $\tilde{V}$ dual to $\vec{V}$ with the fundamental relationship
$$\langle\tilde{V}, \vec{V}\rangle = V_\mu V^\mu = g_{\mu \nu}V^\nu V^\mu= V^2$$
In summary $\vec{V}$ and $\tilde{V}$ are not the same entity since they belong to different vector spaces but they are related via the metric. 
Update:  To emphasize that vectors and one-forms are different geometric objects, consider the following image and caption from the Wikipedia article "One-form"


Linear functionals (1-forms) $\mathrm{α, β}$ and their sum
  $\mathrm{σ}$ and vectors $\mathbf{u, v, w}$, in 3d Euclidean space.
  The number of (1-form) hyperplanes intersected by a vector equals the
  inner product.

A: I would not see it the way @Alfred Centauri has described it. Which might be me misunderstanding the answer/ not understanding the mathematical meaning of different entities here right but I will come back to it after my take on the topic.
There is a physical vector $\mathbf{V}$ and one can express this vector in respect to the  co- or contravariant basis: $$\mathbf{V}=V_\mu\mathbf{e^µ}=V^\mu\mathbf{e_µ}.$$
$\{ \mathbf{e_µ} \}$ and $\{ \mathbf{e^µ} \}$ are just different basis, which are related by $\mathbf{e_µ}\mathbf{e^\nu}=\delta_\mu^{~~\nu}$. The reciprocal basis is not independent of $\{ \mathbf{e_µ} \}$, neither are the resulting components: as they are related by $V_\mu=g_{\mu\nu}V^\nu$. Let me cite the wikipedia page the OP linked on that point:

In a vector space $V$ over a field $K$ with a bilinear form $g : V × V → K$
  (which may be referred to as the metric tensor), there is little
  distinction between covariant and contravariant vectors, because the
  bilinear form allows covectors to be identified with vectors. That is,
  a vector $v$ uniquely determines a covector $\alpha$ via 
  $$\alpha (w)=g(v,w)$$
  for all vectors w. Conversely, each covector $\alpha$ determines a unique
  vector $v$ by this equation. Because of this identification of vectors
  with covectors, one may speak of the covariant components or
  contravariant components of a vector, that is, they are just
  representations of the same vector using reciprocal bases.

I agree with @Alfred Centauri that co- and contravariant vectors and components are not the same but I am not sure about calling them different entities. This might be my mistake because I do not really know what to make of "entities" in a mathematical context but for me it sounds to big of a difference between two so closely related objects.
EDIT: after some points made by @knzhou in the comments and after some additional reading in a modern text book (Wald) on GR (which differs a bit from the "old school" GR lecture notes I was taught GR from).
I think the modern point of view is (as @Alfred Centauri pointed out) to really distinguish between vectors (contravariant) and dual vectors (cotangent, covariant). The equation and points I made above do not distinguish between vectors and dual vectors and I chose a (arbitrary basis/ metric) to make my point. The quote I made actually describes the "intimate" relation between both objects but at a fundamental and basis/metric independent level they are mathematically and geometrically different. There is a relation between them but they are different objects/different entities.
But if one introduces a basis/metric one can use it to

... establish a one-to-one correspondence between vectors and dual
  vectors. Indeed, given a metric $g$ we could use this correspondence
  to entirely circumvent the necessity of introducing dual vectors.
  Normally this is done and accounts for why the concept of dual vectors
  is not more familiar to most physicists. However, in general
  relativity we shall be solving for the metric of the spacetime; since
  the metric is not known from the start, it is essential that we keep
  the distinction between vectors and dual vectors completely clear. [R.M. Wald, 1984, General Relativity, p. 23]

A: You're dealing with different geometric objects: Tangent vectors, which can be realized as equivalence classes of curves, and cotangent vectors, which can be realized as equivalence classes of real-valued functions (think differentials).
There's a natural linear pairing operation between these objects: Compose a curve and a function, and you get a map $\mathbb R\to\mathbb R$. Take it's derivative at the point in question, et voilà. This pairing operation allows us to consider the spaces as 'dual', and in particular identify the cotangent space with the space of linear functionals on the tangent space.
Given a coordinate system on a manifold, the coordinate lines are curves, yielding a basis of the tangent space, whereas the components of the coordinate chart are functions, yielding a basis of the cotangent space. It's easy to show that these bases are algebraically dual, ie their pairing yields the Kronecker delta.
On (pseudo-)Riemannian manifolds, there's additionally a metric tensor $g$, a non-degenerate bilinear form. This tensor induces an isomorphism $g^\flat:v\mapsto g(v,\cdot)$ from the tangent to the cotangent space ('lowering the index'), with an inverse map $g^\sharp$ ('raising the index').
The map $g^\sharp$ can be used to pull back our basis of the cotangent space onto the tangent space, yielding the reciprocal basis. The components of a vector $v$ relative to the reciprocal basis of the tangent space are the same as the components of the covector $g^\flat v$ relative to the dual basis of the cotangent space. This makes it possible to conflate vectors and covectors, but that's considered a mostly bad idea nowadays.
Having said all that, now on to your actual question:

But if I were to take a vector $V^{\mu}$ and lower the index to a covector $V_{\mu}$ in flat space, it most certainly would not be the complicated change of basis matrix shown in the example.  Am I missing something here?

The Minkowski metric is that 'complicated change of basis matrix' - it's just that you're dealing with an orthonormal basis, which makes it simple.
