Are dual quaternions and 3D multivectors isomorphic? Dual quaternions are popularly used as a representation of SE(3) in animation and robotics, with the second 'epsilon' quaternion used to encode translation.  This feels fairly natural, as the fact that rotation can influence position but position cannot influence rotation is built directly into the structure.
Nonetheless, it seems like the 3D multivectors of geometric/clifford algebra might be an even more natural fit, since the units associated with the vector and bivector components would actually match the geometric dimension of the quantities that they encode.  However, when multiplied these allow the odd elements from the two multivectors to combine and feed back into the even 'rotor' components.
Is there a simple mapping between these two algebras, perhaps achieved by pre-multiplying the odd elements by a nilpotent multivector (maybe |1 + e0^e1^e2|), or do these two algebras fundamentally have different behavior?
 A: 
are dual quaternions and 3D multivectors isomorphic?

Not as rings: the first is isomorphic to $\mathbb H[X]/(X^2)$ and the second is isomorphic to $M_2(\mathbb C)$.  The first one has a nonzero central nilpotent element and the second does not. 
As $\mathbb R$ vector spaces, yes, because any two $8$-dimensional real vector spaces are isomorphic.

Would it be fair to say that quaternions are effectively the square roots of scaled rotations?

No, because it suggests “$q^2=kR$ for some rotation $R$ and scalar $k$ is a rotation” which seems meaningless to me.  The fact is that all nonzero quaternions effect rotations using the conjugation action. The ones effecting the same rotation differ by a scalar.
A: $1+e_0e_1e_2$ isn't nilpotent. If your basis vectors square to $-1$ then the algebra is isomorphic to $\mathbb H\oplus\mathbb H$ and there are no nilpotents. If they square to $+1$ then the algebra is isomorphic to $M_2(\mathbb C)$ and there are nilpotents (such as $e_0+e_0e_1$), but there isn't enough room to encode a translation and it certainly isn't natural.
As someone else mentioned, you can introduce an extra dimension with metric signature $0$ to get an algebra isomorphic to the dual quaternions. But they used the full algebra of signature $-{-}0$, which technically is isomorphic but doesn't make physical sense and (as they observed) doesn't have a natural action on vectors. You should instead use signature $-{--}0$ or $+{++}0$ and the even part of the algebra, if you do this. Writing the basis as $\{\hat x,\hat y,\hat z,\hat\epsilon\}$, your general even element can be written $q_1 + \hat x\hat y\hat z\hat\epsilon\,q_2$, where $q_i = a_i \pm (b_i\hat y\hat z + c_i\hat z\hat x + d_i\hat x\hat y)$. (The $\pm$ depends on your signature, and should be chosen to reproduce the standard quaternion convention that $ijk=-1$.)
You can also introduce two new coordinates $\hat t$ and $\hat w$ with $\hat t^2=-\hat w^2$ and only consider terms proportional to $\hat t+\hat w$, but that seems unnecessarily complicated.
Here's a different way of approaching the problem that is perhaps more elegant.
Euclidean 3-space sits somewhat uncomfortably between elliptical and hyperbolic 3-space, both of which have nicer symmetry groups ($O(4)$ and $O(3,1)$ respectively). Given that Euclidean space is an approximation anyway, you may as well pick an approximation that has a nicer symmetry group. For example, you could pick an elliptical space with a curvature radius equal to $R_\text{earth}$. This is not precisely correct, especially in the up-down direction, but neither is x=north-south, y=east-west, z=up-down, and distance=$\sqrt{x^2+y^2+z^2}$. You can represent arbitrary continuous transformations of the elliptical space by elements of the even Clifford algebra of $\mathbb R^4$.
If you implement this, you'll probably run into precision problems because the distances you'll be dealing with will be small compared to $R_\text{earth}$. You'll end up representing positions and transformations as separate large and small parts, scaled by a fixed ratio $\epsilon \sim 1/R_\text{earth}$ in convenient units, and neglecting terms of order $\epsilon^2$. In other words, you'll end up with the dual quaternions again. But this is arguably a nicer way of thinking about them.
A: This probably isn't what you're looking for, but the dual quaternions are isomorphic to a 3D Clifford algebra with a degenerate quadratic form, defined by
$$(xe_1+ye_2+ze_3)^2=-x^2-y^2.$$
$$i=e_1,\quad j=e_2,\quad k=e_1e_2,\quad \varepsilon=e_1e_2e_3$$
The problem here is that, if we're applying this to 3D rigid motions, the "vectors" do not correspond to directions in physical space. So this is no better than the usual "abstract" definition in terms of the mutiplicative relations of $i,j,k,\varepsilon$.
Actually, if we rename $e_3=n_\infty$, then this is clearly related to the conformal geometric algebra of two-dimensional space.
