Restriction on vector fields The 2D vector field (x,-y) does not transform like a vector under rotation(Arfken Vol. 1)! Does this mean we cannot have such a vector field physically?
 A: I'm going to address the important concepts at play here in three dimensions.
The issue here is to get straight the distinction between any function $\mathbf v:\mathbb R^3\to\mathbb R^3$, which we'll call a vector field, and an object that in addition to being a vector field in this sense, transforms in some prescribed way.  To mathematically describe/formalize the latter sort of object, let me first introduce the following definition:  
For any vector field $\mathbf v$, and for any rotation (special orthogonal transformation) $R$, we define a rotated vector field $\mathbf v^R$ by
$$
  \mathbf v^R(\mathbf x) = R \mathbf v(R^{-1} \mathbf x)
$$
This definition is precisely the mathematical formalization of what most people think of when they think of rotating a vector field.
We now define a 3-vector field as a function $\mathbf f(\mathbf v_1, \dots, \mathbf v_n):\mathbb R^3\to\mathbb R^3$ depending on a finite number of vector fields $\mathbf v_i$ such that for any rotation $R$,
$$
  \mathbf f(\mathbf v_1^R, \dots, \mathbf v_n^R) = \mathbf f(\mathbf v_1, \dots, \mathbf v_n)^R
$$
The idea here is that a 3-vector field is a vector field constructed out of a bunch of other vector fields such that if each of the constituent vector fields out of which it is built are rotated, then that is equivalent to just rotating the 3-vector field itself.
As an example, consider the function $\mathbf f(\mathbf{v})$ defined by
$$
  \mathbf f(\mathbf{v}) = \nabla\times\mathbf v
$$
In other words, $\mathbf f(\mathbf{v})$ is just the curl vector field.  Is it a 3-vector field according to the definition above?  Let's check:
\begin{align}
  f(\mathbf{v}^R)(\mathbf x)
&= \nabla\times \mathbf v^R(\mathbf x) \\
&=\epsilon_{ijk}\partial_j(R\mathbf v(R^{-1}\mathbf x))_k \mathbf e_i \\
&= \epsilon_{ijk}\partial_j(R_{kn} v_n(R^{-1}\mathbf x)) \mathbf e_i \\
&= \epsilon_{ijk}R_{kn}\partial_j(v_n(R^{-1}\mathbf x)) \mathbf e_i \\
&=\epsilon_{ijk}R_{kn}(\partial_m v_n)(R^{-1}\mathbf x)\partial_j(R^{-1}\mathbf x)_m \mathbf e_i\\
&= \epsilon_{ijk}R_{kn}(\partial_m v_n)(R^{-1}\mathbf x)\partial_j(R^{-1}_{m\ell} x_\ell) \mathbf e_i\\
&= \epsilon_{ijk}R_{kn}(\partial_m v_n)(R^{-1}\mathbf x) R_{jm}\mathbf e_i \\
&= R_{i\ell}\epsilon_{\ell m n}(\partial_m v_n)(R^{-1}\mathbf x)\mathbf e_i \\
&= R(\nabla\times \mathbf v)(R^{-1}\mathbf x) \\
&= f(\mathbf v)^R(\mathbf x)
\end{align}
or in summary
$$
  \mathbf f(\mathbf v^R) = \mathbf f(\mathbf v)^R
$$
so, indeed, the curl of a vector field is a 3-vector field!  Now, let's return to your example.  Let's take a 3D analog and define a function $\mathbf f(\mathbf v)$ by
$$
  \mathbf f(\mathbf v)(\mathbf x) = (v_1(\mathbf x), -v_2(\mathbf x), v_3(\mathbf x))
$$
Is this a 3-vector field?  Namely, if we invert only one of the coordinates, then is the resulting vector field a 3-vector field according to our definition?  I claim that no, it is not.  To see this, consider the rotation $R$ that leaves the $z$ axis fixed, but that rotates in the $x$-$y$ plane by $\pi/2$ clockwise.  I'll leave it to you to check that
$$
  \mathbf f(\mathbf v^R)(x,y,z) = (-v_2(y,-x,z),-v_1(y, -x, z), v_3(y, -x, z))
$$
while
$$
  \mathbf f(\mathbf v)^R(x,y,z) = (v_2(y,-x,z),v_1(y, -x, z), v_3(y, -x, z))
$$
so that for this rotation $R$,
$$
  \mathbf f(\mathbf v_R)\neq \mathbf f(\mathbf v)^R
$$
and therefore this particular $\mathbf f(\mathbf v)$ is not a 3-vector!
I think this is rigorously the meaning of what Arfken is trying to say.
Physical Interpretations
You might ask why the definition of a 3-vector field given above is useful in physics.  Well here's the main idea.  Let's say that we measure a certain physical vector field using some apparatus, like the velocity field $\mathbf v$ on the surface of a lake for instance.  Then given this velocity field, one could compute the curl of this field $\mathrm{curl}(\mathbf v)$.  Now suppose that we were to rotate our measuring apparatus, then we would measure the rotated velocity field $\mathbf v^R$.  We could now calculate the curl $\mathrm{curl}(\mathbf v^R)$ of this rotated vector field, but because, as we showed above, the curl of a vector field is a 3-vector field, we could just as easily compute the rotated curl $\mathrm{curl} (\mathbf v)^R$, and we would get the same answer.  So in some sense, a 3-vector can be interpreted as a computed physical quantity which rotates in the same way as measured physical quantities.
A: Transforming the positions and rotating the vector field are two different things.
Let $p$ be a point--for instance, $p = x e_1 + y e_2$, with $e_1, e_2$ being the usual Cartesian basis vectors.  We can define a transformation $f(p) = p' = x' e_1 + y' e_2$.
Now then, let $A(p)$ be a vector field--in our case, $A(p) = x e_1 - y e_2$.  As a vector field, $A$ can be considered as the derivative of some curve through the point $p$.  Let $c(\tau,p)$ be such a curve.  Then $A = \partial c/\partial \tau$ for some curve $c$.  There are necessarily different curves through each point $p$, but such curves always exist.
It is this construction that is crucial to understanding the transformation properties of vector fields.  Let us now consider $c' = f \circ c$, which is a family of curves in the primed space.  The chain rule then tells us that
$$\frac{\partial c'}{\partial \tau} \Big|_{p'} = \frac{\partial c}{\partial \tau} \cdot \nabla f \Big|_p$$
The object $a \cdot \nabla f$ for any vector $a$ is special--we call it the "differential" of $f$, or the "Jacobian".  Denote this as $\underline f(a)$, and we get
$$\frac{\partial c'}{\partial \tau} \Bigg|_{p'} = \underline f \left[\frac{\partial c}{\partial \tau} \right]_{p}$$
Or, more concisely,
$$A'(p') \equiv \underline f(A[p])$$
Linear transformations are special--they obey $f = \underline f$.  This means that $A'$ is necessarily just the rotation of $A$, but evaluated at the rotated point also.
Let's work out what this means for your vector field.  First work out the rotation:
$$\underline f(e_1) = e_1 \cos \theta + e_2 \sin \theta \\
\underline f(e_2) = e_2 \cos \theta - e_1 \sin \theta$$
Now work through the vector field:
$$\begin{align*} A(p) &= x e_1 - y e_2 \\
\underline f(A[p]) &= (x \cos \theta + y \sin \theta) e_1 + (x \sin \theta - y \cos \theta) e_2
\end{align*}$$
We need to convert this into the primed coordinates.  See that
$$x = x' \cos \theta + y' \sin \theta \\
y = y' \cos \theta - x' \sin \theta$$
Let's look at the first component.
$$x \cos \theta + y \sin \theta = x' \cos^2 \theta +2 y' \sin \theta \cos \theta - x' \sin^2 \theta = x' \cos 2\theta + y' \sin 2\theta$$
Similar logic applies for the other component, yielding
$$A'(p') = (x' \cos 2 \theta + y' \sin 2 \theta) e_1 + (x' \sin 2\theta - y' \cos 2\theta)e_2$$
You may be skeptical that this is the correct transformation of the vector field, but I assure you it is.  To check, pick the point $p = e_1$ and a vector $v = e_2$.  We know that $A(e_1) = e_1$, so that $v \cdot A = 0$.  If we transform both vectors according to $\underline f$, we should get $\underline f(v) \cdot \underline f(A) = 0$ as well--after all, a rotation should not change angles, should not change orthogonality.
Of course, it would be onerous to find the coordinates of $p = e_1$ in the primed frame.  It's much easier to see that
$$A'(p') = e_1 \cos \theta + e_2 \sin \theta$$
Similarly, $v' = \underline f(v) = e_2 \cos \theta - e_1 \sin \theta$.  This does indeed satisfy that $v' \cdot A' = 0$, as required.
In short, you cannot merely rotate the vector itself.  There are two rotations involved: one of the underlying positions, and then one of the vector field.  You should not expect that $A' = x' e_1 - y' e_2$.  This is not founded in the transformation law.  Perhaps this is what Arfken meant--that you can't be naive, expecting that even a simple rotation will preserve components of vectors and their relations to positions.  Once you derive the transformation law for vectors, though, it becomes a bit meaningless to say something is or is not a vector field. It's easy enough to impose this transformation law.  Perhaps Arfken meant to say that, given $A$ and $A'$ as prescribed in your question, it's clear that is inconsistent with the vector field transformation law.  Really, though, this strikes me as quite unclear.
