I want to know why there is no solution $A_{n\times n}$ for $A^2 = -I$ , when the dimension of A is odd. The question is to find a matrix $A_{n\times n}$ for $A^2 = -I$. In geometric, I can imagine that A is a rotation transformation(rotate by $\frac{\pi}{2}$), and for even number n, that can be popularized by block matrix.(Then I found the solution for even number n)
However, I can't verify why there is no solution for odd n.
To summarize, I want know why there is no solution for $A^2 = - I$ , when square matrix A is  odd demension.
 A: The answer by F_M_ is clear and easy, but I wanted to type something that is closer to what you already tried and more geometric.
We proceed by induction on $n$. For $n = 1$ we are just looking at the fact that $a^2 = -1$ has no (real) solutions.
For $n > 1$ odd, just pick a vector $v_1$ and define $v_2 := Av_1$.
$v_1$ and $v_2$ span a two-dimensional plane $P$. By definition $Av_1 \in P$ but since $Av_2 = -v_1$ by the nature of $A$ we find that $Av_2 \in P$ as well and hence every point in $P$ is mapped to a point in $P$ under $A$.
Restricting our attention to what happens in the two-dimensional plane $P$ it is also a bit easier to think about $A$ geometrically. It could be a quarter turn rotation, as you write, but also a quarter turn rotation combined with some stretching and shrinking, e.g.
$\begin{pmatrix} 0 & -3 
\\ 1/3 & 0 \end{pmatrix}$.
There is however a way in which we can think about $A$ as just being an ordinary quarter turn rotation and in fact this is very useful in the proof!
This is a trick you should have seen once in your life because it is not so easy to come up with from scratch but very useful.
Let $( , )$ be the ordinary inner product on $\mathbb{R}^n$. We define a new inner product $\langle , \rangle$ by
$$\langle v, w \rangle = 1/4((v, w) + (Av, Aw) + (A^2v, A^2w) + (A^3v, A^3w))$$
Making use of the fact that $A^4 = I, A^5 = A$ etc we see that
$$\langle Av, Aw \rangle = \langle v, w \rangle$$  for all  $v, w \in \mathbb{R}^n$.
In other words: if we measure lengths and angles by this new inner product instead of the old one we find that $A$ is a rigid, distance and angle preserving motion. In particular its action in the plane $P$ is now a quarter turn rotation.
But what is more interesting is what happens in the space $P^\perp$ of vectors that are orthogonal to all vectors in the plane $P$ under this new inner product. For each $w \in P^\perp$ and $p \in P$ we have that $\langle w, p \rangle = 0$ and hence that $\langle Aw, Ap \rangle = 0$. But since every point in $P$ is of the form $Ap$ for some other point $p \in P$, this means that $Aw \in P^\perp$ as well!
In other words: $A$ maps points in $P^\perp$ to points in $P^\perp$ and hence $P^\perp$ is an $(n-2)$-dimensional, hence odd-dimenional space of dimension $< n$ on which $A$ acts. Thus we can restrict our attention to $P^\perp$, apply the induction hypothesis and conclude that $A$ does not exist there. But that means it didn't exist on the big space $\mathbb{R}^n$ in the first place either!
A: Since $n$ id odd, characteristic polynomial of $A$ has a real root (i.e. any odd degree polynomial has a real root). Let $\lambda$ be eigenvalue and $x$ corresponding eigenvector (nonzero). Then, since $Ax = {\lambda}x$ we get $A^2x = {\lambda}^2x$.
At the same time, if $A^2 = -I$ then $A^2x = -x$ so, for a real number ${\lambda}^2 = -1$
