What is the redshifted amplitude of a gravitational wave? Consider a gravitational plane wave in flat background spacetime, with amplitude $h$ and frequency $f$. For an observer moving with redshift $(1+z)$ relative to the plane wave, what is the observed amplitude?
 A: GR tells us that a gravitational wave (GW) does the transformation called strain (=fractional change of length) as the wave passes.  Rotation, boost, and strain transformations belong to the group SL(4).  Rotations and boosts form the Lorentz Group which is a subgroup of SL(4).  First, we extend the Lorentz Group to all the transformations of SL(4) by specifying the rest of the parameters beyond $\vec{\theta} \ and \ \vec{\lambda}$, and describe what the transformations do.  Then we can simply answer what a boosted observer sees a GW do.
\begin{align}
\Theta_j^{\ \ \ i} & =-\frac{1}{2}\begin{bmatrix}
         0 & \theta_3 &-\theta_2 & \mu_1 \cr
 -\theta_3 &        0 & \theta_1 & \mu_2 \cr
  \theta_2 &-\theta_1 &        0 & \mu_3 \cr
    -\mu_1 &   -\mu_2 &   -\mu_3 &     0 \end{bmatrix} _{Asym} + 
\frac{1}{2}\begin{bmatrix}
   h_{+1} &    h_{X3} & h_{X2}    & \lambda_1 \cr
   h_{X3} &    h_{+2} & h_{X1}    & \lambda_2 \cr
   h_{X2} &    h_{X1} & h_{+3}    & \lambda_3 \cr
\lambda_1 & \lambda_2 & \lambda_3 &    h_{+4} \end{bmatrix} _{Sym}\\ 
\\
Group\quad element & =e^{\Theta_j^{\ i}J_i^{\ j}} \quad where \quad [J_i^{\ j},J_k{\ ^l}]=(\delta_i^{\ l} J_k^{\ j}-\delta_k^{\ j} J_i^{\ l})
\quad\quad i,j=1...4 \\
det(e^{\Theta_j^{\ i}J_i^{\ j}}) & =1 \quad \Rightarrow \quad Trace(\Theta)=0=h_{+1}+h_{+2}+h_{+3}+h_{+4} \end{align}
In the 4x4 representation each generator $J_i^{\ j}$ is a 4x4 matrix with a 1 in (row i, column j). Then the group element becomes a 4x4 matrix $M=e^{\Theta}$.
The co-variant (down) and contr-varient (up) indices on $\Theta$ and $J$ indicate how $\Theta$ and $J$ transform when viewed from another frame.
$$
\Theta'=M\Theta M^{-1} \quad J'=MJM^{-1}
$$
$\vec\theta=\theta \hat{n}_\theta \quad$  where $\theta$ is the rotation angle (in radians) about the unit vector $\hat{n}_\theta$ (right hand rule).
$\vec\lambda=\lambda \hat{n}_v \quad$ where $\lambda=\tanh^{-1}(v/c)$ is the boost (in radians) in the direction of the velocity unit vector $\hat{n}_v$.
$\vec\mu \quad$ are rotation angles (in radians) of a 4-vector in a space-time plane. For example, $$ x'=\cos{\mu_1}\ x+\sin{\mu_1}\ ct \\ ct'= -\sin{\mu_1}\ x + \cos{\mu_1}\ ct$$  We have not dealt with space-time rotations before in physics.  The transformations are not part of the Lorentz group and do not leave $x^2-(ct)^2$ invariant.
$h_{...}\quad$ these are the space-space strains and time-time strain (in radians). They have been subscripted the same as they are for gravitational waves. For example, $h_{+1}$ stretches an object in the x direction, and $h_{X1}$ parallelepipeds an object in the x-y plane.  These are the two "gravitational wave polarizations" for a GW travelling in the z direction.
Notice that because the boosts $\vec{\lambda}$ belong to the symmetric part of $\Theta$, they are strains (not rotations).  That's why expanding
$$
M=e^{\Theta}=I+\Theta+\dfrac{\Theta^2}{2!}+ \dfrac{\Theta^3}{3!}+ \dfrac{\Theta^4}{4!} +…
$$ 
for boosts yields cosh's and sinh's rather than cos's and sin's. The "old time" physicists complexified the algebra and introduced the imaginary angles $i\vec{\lambda}$ so they could continue to do orthogonal group rotations with sin's and cos's of imaginary angles.  It would be unnecessarily confusing to continue this historical artifact here.
All of this probably unfamiliar prelude was to be able to answer the question: "If a GW does a strain (for example on LIGO), what does a boosted observer see?"  This can now be answered by just transforming $\Theta$ by a boost $\lambda$ to get $\Theta'$.
The question asks for a boost in the same direction as the GW is travelling. So, assume both are in the z direction, and therefore the GW strains are in the x-y plane:
\begin{align}
\Theta{'}  & = \begin{bmatrix}  
1 & 0 &              0 &              0 \cr
0 & 1 &              0 &              0 \cr
0 & 0 & \cosh{\lambda} &-\sinh{\lambda} \cr
0 & 0 &-\sinh{\lambda} & \cosh{\lambda} \end{bmatrix} 
\begin{bmatrix}
h_+ &  h_X & 0 & 0 \cr
h_X & -h_+ & 0 & 0 \cr
  0 &    0 & 0 & 0 \cr
  0 &    0 & 0 & 0 \end{bmatrix} 
\begin{bmatrix}  
1 & 0 &              0 &              0 \cr
0 & 1 &              0 &              0 \cr
0 & 0 & \cosh{\lambda} & \sinh{\lambda} \cr
0 & 0 & \sinh{\lambda} & \cosh{\lambda} \end{bmatrix}\\
&=\begin{bmatrix}
h_+ &  h_X & 0 & 0 \cr
h_X & -h_+ & 0 & 0 \cr
  0 &    0 & 0 & 0 \cr
  0 &    0 & 0 & 0 \end{bmatrix} 
\end{align}
Notice that the strain amplitudes are unchanged! Therefore, the answer to the original question is that the GW amplitudes appear unchanged to an observer boosted in the same direction the GW is travelling. 
This is completely different from how the transverse electric and magnetic fields of an EM wave transform. We see this by transforming the EM field tensor F of $\vec{E},\vec{B}$ to a frame boosted in the z direction.  The EM wave is going in the z direction so $E_z=B_z=0$.
\begin{align}
F'  & = \begin{bmatrix}  
1 & 0 &              0 &              0 \cr
0 & 1 &              0 &              0 \cr
0 & 0 & \cosh{\lambda} &-\sinh{\lambda} \cr
0 & 0 &-\sinh{\lambda} & \cosh{\lambda} \end{bmatrix} 
\begin{bmatrix}
   0 &     0 & -cB_y & E_x \cr
   0 &     0 &  cB_x & E_y \cr
cB_y & -cB_x &     0 &   0 \cr
 E_x &   E_y &     0 &   0 \end{bmatrix} 
\begin{bmatrix}  
1 & 0 &              0 &              0 \cr
0 & 1 &              0 &              0 \cr
0 & 0 & \cosh{\lambda} & \sinh{\lambda} \cr
0 & 0 & \sinh{\lambda} & \cosh{\lambda} \end{bmatrix}\\
&=\begin{bmatrix}
              0 &  0 &-\gamma(cB_y-\beta E_x) &\gamma(E_x-\beta cB_y) \cr
              0 &  0 & \gamma(cB_x+\beta E_y) &\gamma(E_y+\beta cB_x) \cr
\gamma(cB_y -\beta E_x) &-\gamma(cB_x+\beta E_y) &               0 & 0\cr
\gamma(E_x -\beta cB_y) & \gamma(E_y+\beta cB_x) &0 &  0 \end{bmatrix}\end{align}
where the substitutions $\gamma=\cosh{\lambda}$ and $\beta\gamma=\sinh{\lambda}$ were made.  
For the boost being in the same direction as the GW is travelling, the frequency of the GW seen by the boosted observer is redshifted just like for an EM wave.
$$
\nu’=\frac{\nu}{1+z}=e^{-\lambda}\nu
$$
