Linear transformation on $\mathbb{R}^7$ Consider the linear transformation on $T : \mathbb{R}^7 \rightarrow \mathbb{R}^7$ defined by $$ T(x_1,x_2,...,x_7)=(x_7,x_6,...,x_1)$$ Then which of the following are true ?

1) $det(T)=1$
2) $\exists$ a basis $\mathcal{B}$ of $\mathbb{R}^7$ such that $[T]_\mathcal{B}$ is diagonal
3) $T^7=I$
4) the smallest $n$ such that $T^n=I$ is even.

Mt attempt:
A simple calculation shows $T^2=I$,  so 4) is true and 3) is false
For 1):  In general, the determinant of $T= (-1)^{\text{dim V}}.(\text{constant term of the characteristic polynomial of T) }$
So, we must find the matrix of $T$ in any basis and hence we find the determinant of that matrix, which is same as $det(T)$
My Question is:
Is there any short cut way to find the determinant of T ?
How to approach 2) ?
I'm  find the matrix of $T$ w.t.to the standard basis. But that matrix is not diagonal.
 A: It takes three transpositions of the seven standard unit vectors of $\mathbb{R}^7$ to create $T$: $\vec{e}_1\leftrightarrow\vec{e}_7$, $\vec{e}_2\leftrightarrow\vec{e}_6$, $\vec{e}_3\leftrightarrow\vec{e}_5$. So $T$ is an odd permutation of the seven standard unit vectors of $\mathbb{R}^7$. So it's determinant is $(-1)^3=-1$.
$T$'s matrix with respect to the standard basis is antidiagonal with all $1$s. So $T$'s standard matrix is a symmetric matrix. So $T$ is (orthogonally) diagonalizable. That makes item 2) true.
In fact, it's easy to see what the basis $\mathcal{B}$ could be. It's clear that $\vec{v}_4$ is one eigenvector of eignevalue $1$. So are $\vec{v}_1+\vec{v}_7$, $\vec{v}_2+\vec{v}_6$, and $\vec{v}_3+\vec{v}_5$. And some eigenvectors with eigenvalue $-1$ are $\vec{v}_1-\vec{v}_7$, $\vec{v}_2-\vec{v}_6$, and $\vec{v}_3-\vec{v}_5$. These are seven linearly independent eignevectors.
A: The given linear transformation takes the basis (in the given order) and reverses the order.  The reversal permutation of any number of items is achieved by exchanging the extreme elements, then again exchanging the extremes in the remaining elements and so on. (if odd number of elements, the middle one left alone will be fixed).
Now look at the alternative basis: $e_1+e_7, e_1-e_7, e_2+e_6, e_2-e_6, e_3+e_5, e_3-e_5, e_4$. Easy to verify that they generate all the $e_i$'s and are in the right number hence a basis.
We can check that $T$'s effect on this new basis: they are fixed or sent to their negatives. These are the eigenvectors forming a basis, providing diagonalization. SO determinant is computed immediately.
A: If $A$ is the permutation matrix corresponding to the permutation $\rho$
then $\det A$ is the sign of $\rho$.
If $T^2=I$, what does that tell you about the eigenvalues of $T$?
