How does the Lorentz transformation $\Lambda^{\mu}{}_{\nu}$ transform? For example the Four-velocity transforms as $$U^{a'}=\Lambda^{a'}{}_{\nu}U^{\nu},$$
the Faradaytensor as $$F^{a'b'}=\Lambda_{\,\,\mu}^{a'}\Lambda_{\,\,\nu}^{b'}F^{\mu\nu}$$ or in Matrixnotation:
$$F'=\Lambda F\Lambda^{T},$$
where $\Lambda^{T}$ is the Transpose of the Matrix.
But the Lorentz matrix $\Lambda^{\mu}{}_{\nu}$ is not a tensor.
Does $\Lambda$ transform anyway like a second rank tensor in the same way like the Faradaytensor?
 A: Indeed, to add to orion's answer, the definition of a vector (and / or spinor) in physics (as opposed to the mathematical definition as an element of a linear space over a field) is most often stated in terms of how the object concerned transforms (e.g. see my answer here) when "co-ordinate transformations" are made: more precisely, when one switches between two overlapping charts of a manifold. Vectors transform like the members of tangent spaces to manifolds. "Covectors" or one-forms are linear functionals of vectors. Then tensors are simply general multilinear functionals of vectors and / or one-forms. I really like the language and teaching style of Misner Thorne and Wheeler "Gravitation" here, also replicated in the early part of Kip Thorne's lectures here.
The Lorentz transformation, on the other hand is a kind of co-ordinate transformation, and, as such, a vector / oneform / tensor must, by definition, transform in the prescribed way by it.
So tensors, vectors and n-forms are defined by how their components behave in response to co-ordinate transformations. If you like, tensors, vectors and n-forms are a kind of "software" (or "machine" to use Kip Thorne's and Misner/Thorne/Wheeler wording) that is built to a specification of how that software/machine must react to various inputs. In this analogy, the Lorentz transformation is one of the inputs for the machine addressed by the specification.
A: It is a transform, not a tensor. Tensors describe a physical quantity at a selected point in time-space and have to transform accordingly. But the Lorentz matrix doesn't describe any physical quantity in a single frame, it's just a change of variables between two coordinate frames. You can understand this by analogy with 3D rotations. Transforms are composed (multiplied) together: if you go from coordinates $a$ to $a'$ and then from $a'$ to $a''$, you multiply the transforms to get a direct transform from $a$ to $a''$.
A: The Lorentz transformation does not transform, since it is not an object living on the spacetime manifold in the way that vectors and tensors do.
In general, the objects that we think of as "vectors" or "tensors" are elements of the tensorial powers of the (co)tangent spaces at every point of the spacetime manifold. Under any coordinate transformation $x^\mu \mapsto y^\mu(x)$, an element of the tangent space will transform as 
$$v^\mu \mapsto \frac{\partial y^\mu}{\partial x^\nu}v^\nu$$
while an element of the cotangent space transforms as 
$$ v_\mu \mapsto \frac{\partial{x^\nu}}{\partial y^\mu}v_\nu$$
These extend by linearity to arbitrary tensor products of the (co)tangent spaces. Now, a Lorentz transformation is just a coordinate transformation such that $x \mapsto \Lambda x$, so that the derivatives acting on the (co)tangent spaces are $\Lambda$ and $\Lambda^{-1}$ at every point, respectively.
The $\Lambda$ is not a member of any (co)tangent space, it does not belong to a particular point, but is the derivative of the general coordinate transformation applied.
A: Since, to take a specific example; for two $4$-vectors $a^{\mu}, b^{\mu}$ it's not too difficult to show that the scalar product
\begin{eqnarray}
a \cdot b &=& a_{\mu}b^{\mu} \\
          &=& \eta_{\mu \nu}a^{\mu}b^{\nu} 
\end{eqnarray}
is invariant under boosts. Now, what are the transformations that leave this scalar product (and by implication the metric) invariant? Examine this scalar product
\begin{equation}
a' \cdot b' = (\Lambda^{\mu}_{\lambda}a^{\lambda})\eta_{\mu \nu}(\Lambda^{\nu}_{\rho}b^{\rho})
\end{equation}
Clearly this works for all matrices $\Lambda$ such that
\begin{equation}
\Lambda^{\mu}_{\lambda}\Lambda^{\nu}_{\rho}\eta_{\mu \nu} = \eta_{\lambda \rho}
\end{equation}
Celarly this is just $\Lambda^{T} \eta \Lambda$ in matric notation.
