A is involutory matrix so is every natural power of A If A is involutory matrix then every natural power of A is involutory. Its written on Wikipedia 
I didn't get it. Suppose $A^7=A^6.A \ne I$.
So how come A is involutory ?
https://en.wikipedia.org/wiki/Involutory_matrix Fourth property.
 A: In the article, there is only one direction specified:

If $\mathbf{A}$ is involutory, then $\forall n\in\mathbb{N}:\mathbf{A}^n$ is involutory, and 
  
  
*
  
*if $n$ is odd, then $\mathbf{A}^n=\mathbf{A}$
  
*if $n$ is even, then $\mathbf{A}^n=\mathbf{I}$

In other words, every power is again involutory, as they relate to either $\mathbf{A}$ or $\mathbf{I}$ and $\mathbf{A}$ was supposed to and $\mathbf{I}$, the identity, trivially is.
Where your confusion comes from, is that you've read the statement as the converse implication, i.e. that if $\mathbf{A}^n$ is involutory, then $\mathbf{A}$ is which is generally not true.
EDIT: I give a proof the above mentioned statement: Let $\mathbf{A}$ be an involution, i.e. $\mathbf{A}^2=\mathbf{I}$. Let's look a the power $\mathbf{A}^n$ for $n\in\mathbb{N}$.
Let $n$ be even, i.e. $n=2k$. Now, we can write $\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2$($k$ times). But now, $$\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2=\mathbf{I}\dots\mathbf{I}=\mathbf{I}$$ For $n$ odd, i.e. $n=2k+1$, we may write $\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2\mathbf{A}$ with k many $\mathbf{A}^2$. Now, $$\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2\mathbf{A}=\mathbf{I}\dots\mathbf{I}\mathbf{A}=\mathbf{A}$$.
As $\mathbf{A}$ and $\mathbf{I}$ are involutory and 
either$\mathbf{A}^n=\mathbf{A}$ or $\mathbf{A}^n=\mathbf{I}$, we have that $\mathbf{A}^n$ with the desired equalities shown before.
A: A matrix $A$ is involutory if $A^2=I$. In particular $A$ is invertible.
An elementary property of powers is that $(A^m)^n=A^{mn}=(A^n)^m$. If $A$ is involutory, then
$$
(A^n)^2=(A^2)^n=I^n=I
$$
so $A^n$ is involutory as well.
Actually, if $n=2k+1$ is odd, then $A^n=A^{2k+1}=A^{2k}A=IA=A$; if $n=2k$ is even, then $A^n=A^{2k}=I$.

If $A^7=A^6$ and $A$ is invertible, then $A=I$.
