bit reversed order of FFT matrix for channel estimation
So, writing your $F^b$ and $F^{'b}$ as permutations $p, p^{-1}$ of the DFT matrix $F$ , IDFT $F'$, and the cyclic prefixing and removal also as Matrix operation $P$, $R$ (which boils down to cyclically extended or truncated identity matrices), and the circular convolution with the channel impulse response as circulant matrix $C$,
\begin{align}
x&amp;=F^{'b}X=p F' X\\
x_{cp} &amp;= PpF' X\\
y_{cp} &amp;= y*x_{cp} \\
y&amp;=R(y*x_{cp})= CpF'X\\
Y&amp;=F^By \\
&amp;= p'FCpF'X &amp; F, F' \text{ are symmetric matrices!}\\
&amp;=p'FCF'pX\\
&amp;=p'\underbrace{FCF'}_{\text{IDFT-Channel-DFT}}\underbrace{pX}_{\text{permuted symbols}}
\end{align}
So, all that's really happening here mathematically is that you permute the symbols before transmission, then de-permute them at the receiver. In between, you get normal OFDM. There's no advantage or disadvantage to the permutation w.r.t. to channel diagonalizability, but we've arrived here in your previous question already.
Now, for channel estimation, you'll want to insert pilots such that they make sense for the central $FCF'$ part – in other words, you need to think about where you'd want your normal OFDM system to have these pilots, and then shift these positions according to your $p$.
Of course, the channel you estimate left of your final reordering isn't the same as the channel you estimate with OFDM, it's a permutation. As shown above, it's however &quot;just&quot; a permutation, there's nothing in the maths here that changes anything about the channels properties at all, so if your results aren't the same as for straightforward OFDM, then you've got an implementation bug somewhere.