Here I try to derive only the mutual information. As I discuss later, the problem seems to be difficult in general.
To obtain capacity, the important point is to characterize the conditional probability $P_{Y|X}$ of the channel. After that $H(Y)$ and $H(Y|X)$ can be computed as a function of the input distribution $P_X$ which should be optimized later to maximize $I(X;Y)$ and obtain the capacity.
Consider a fixed $n$ and let $X$ be a column vector of dimension $n$. Suppose the the entries of $X$ are discrete so we use discrete entropy. The channel can be written as:
$$
Y=AX,
$$
where $A$ is randomly chosen from the following subset of permutation matrices:
$$
\mathcal A=\left\{A\in\mathbb R^{n\times n}: A=\begin{pmatrix}I_1&0&0\\0&J_2&0\\0&0&I_3\end{pmatrix}\text{ for some } I_1,I_3,J_2\right\},
$$
with $I_1$ and $I_3$ identity matrices and
$$
J_2=\begin{pmatrix}0&\dots&0&1\\0&\dots&1&0\\1&\dots&0&0\end{pmatrix}.
$$
Note that $J_2$ acts on the block that is reversed. The matrix $A$ is randomly generated by randomly picking a block and associating $J_2$ to that block.
If $Y=y$ is a block-reversed version of $X=x$, then the matrix $A$ can be determined from $x$ and $y$. The matrix is not unique however. Denote the set of these matrices by $A(x,y)$. If $y$ is a block reversed version of $x$, we have:
$$
P(Y=y|X=x)=P(AX=y|X=x)=P(A\in A(x,y))
$$
Otherwise $P(Y=y|X=x)=0$. Define the set $T$ and $T_y$ as follows:
$$
T=\left\{(x,y):Ax=y\text{ for some } A\in\mathcal A\right\}\\
T_y=\left\{x: Ax=y\text{ for some } A\in\mathcal A\right\}.
$$
Hence if $(x,y)\in T$, then $y$ is a block-reversed version of $x$.
Since no particular assumption is given here, we examine uniform distribution as an example.
Assume that we have a uniform distribution on $A$ with $P(A=A_0)=\frac 1{\binom{n}{2}}$. Then, this implies that:
$$
P(Y=y|X=x)=\frac {|A(x,y)|}{\binom{n}{2}}\\
P(Y=y)=\sum_{x}P(Y=y|X=x)P(X=x)=\sum_{x\in T_y}\frac{|A(x,y)|}{\binom{n}{2}}P(X=x)=\frac{E(|A(X,y)|)}{\binom{n}{2}}.
$$
Therefore:
\begin{align}
H(Y|X)&=E(\log\frac{1}{P(Y|X)})\\
&=\sum_{(x,y)\in T}P(X=x)\frac {|A(x,y)|}{\binom{n}{2}}\log{\frac {\binom{n}{2}}{|A(x,y)|}}\\
&=\frac{\log{\binom{n}{2}}}{\binom{n}2}\sum_{y}E(|A(X,y)|)+\sum_{(x,y)\in T}P(X=x)\frac {|A(x,y)|}{\binom{n}{2}}\log{\frac {1}{|A(x,y)|}}.
\end{align}
and
\begin{align}
H(Y)&=E(\log\frac{1}{P(Y)})\\
&=\sum_{y}E(|A(X,y)|)\frac{1}{\binom n2}\log\frac{{\binom{n}{2}}}{E(|A(X,y)|)}
\end{align}
So the mutual information is obtained as:
$$
I(X;Y)=\frac{1}{\binom n2}\sum_{(x,y)\in T}P(X=x)|A(x,y)|\log\frac{|A(x,y)|}{E(|A(X,y)|)}.
$$
I am not sure how useful this expression is. From now on, we require to know more about the support of $X$. Even with that, e.g. assume vectors on $\mathbb F_2^n$, I do not see any straightforward way to obtain the optimal distribution.