Questions on transposing in the opposite diagonal I am working through some problems on a function that acts as a transposition but in the opposite diagonal to the main diagonal, denoted $A^r$ for $A\in M_{n\times n}(\mathbb{R})$ but I am unsure of workings and stuck on the final parts.

(1) What is the $ij$-entry of $A^r$? Show that $(AB)^r=B^rA^r$ and $\text{det}(A^r)=\text{det}\,A$.

I can write the operation of $A^r$ as reversing the order of elements in each row, then applying a regular transposition, and then reversing the order of elements in each row again. This gives
$$\large{a_{ij}\mapsto a_{i,n+1-j}\mapsto a_{n+1-j,i}\mapsto a_{n+1-j,n+1-i}}.$$
Now
$(AB)_{ij}=\sum_{k=1}^na_{ik}b_{kj}$ gives $((AB)^r)_{ij}=\sum_{k=1}^na_{n+1-j,k}b_{k,n+1-i}$, and $B^rA^r$ has $ij$-entry
$${\sum_{k=1}^nb_{n+1-k,n+1-i}a_{n+1-j,n+1-k}}=\sum_{k=1}^na_{n+1-j,k}b_{k,n+1-i}=((AB)^r)_{ij},$$
by mapping dummy variable $k\mapsto n-k$.
To show $\text{det}(A^r)=\text{det}A$: Let $\bar{A}$ denote the resulting matrix after reversing the order of the elements in each row $A$. This can be achieved through swapping columns a number of times $t_n$ that is fixed for a given $n$ (for $M_{n\times n}(\mathbb{R}))$, so we can say $\text{det}(\bar{A})=(-1)^{t_n}\text{det}A$. So noting $A^r$=$\bar{((\bar{A})^T)}$,
$$\text{det}A^r=\text{det}\bar{((\bar{A})^T)}=(-1)^{t_n}\text{det}\bar{A}^T
=(-1)^{t_n}\text{det}\bar{A}=(-1)^{2t_n}\text{det}A=\text{det}A.$$
I am pretty sure this solution is correct, but maybe there is a simpler way? I'm not sure if this answer is more complicated than it needs to be.

(2) Show that $\phi:GL_n(\mathbb{R})\rightarrow GL_n(\mathbb{R})$ defined by $\phi(A)=(A^r)^T$ is a well-defined group isomorphism with $\phi^2$ being the identity. What the image of $\phi$ restricted to the upper triangular matrices?

I am happy with these parts , and have that the required image is the lower triangular matrices. But I am stuck on the next part:

(3) Is there a $C\in GL_n(\mathbb{R})$ such that $\phi(A)=C^{-1}AC$ $\forall\,A\in GL_n(\mathbb{R})$?


(4) Let $A$ be a matrix of a linear transformation $T$ with respect to $(v_1,...,v_n)$. Find a basis such that $\phi(A)$ is the matrix of $T$.

For (3) I can see that if such a $C$ exists, then we find $(C^r)^T=C$ and $((C^{-1})^r)^T=C$ which restricts $C$ and its inverse to a kind of symmetry where
$$\Large{c_{ij}=c_{n+1-i,n+1-j}}$$
but I am unsure how to proceed. Any hints would be great!
 A: A certain number of operations can be made very simple by the existence of a closed form formula for $A^r$:
$$A^r=JA^TJ\tag{1}$$
where
$$J:=\begin{pmatrix}0&0&\cdots&0&1\\
0&0&\cdots&1&0\\
&&\cdots&&\\
0&1&\cdots&0&0\\
1&0&\cdots&0&0\end{pmatrix}$$
with the property that $J^T=J=J^{-1}$.
Proof of (1): $J$ is a permutation matrix. The RHS of formula (1) amounts to the composition of 3 symmetries:

*

*post-multiplication by J reverses the order of columns (symmetry wrt $y$ axis)


*transposition is a symmetry with respect to straight line with equation $y=-x$.


*pre-multiplication by $J$ reverses the order of rows (symmetry wrt $x$ axis).
The composition of these 3 symmetries is a symmetry with respect to line with equation $y=x$ as can be seen for the image of an orthogonal basis.
Related: the answers to this question.

Using formula (1), formulas
$$(AB)^r=B^rA^r, \ \ \det(A^r)=\det(A), etc.$$
are very easy to establish.
Let us prove Question (3) affirmatively using (1):
There exists $D$ such that
$$A^T=D^{-1}AD\tag{2}$$
Indeed, any matrix is conjugated with its transpose.
Plugging (2) into (1) gives
$$A^r=J^{-1}D^{-1}ADJ=(\underbrace{DJ}_C)^{-1}A(\underbrace{DJ}_C)$$
