How valid is this concept or does this already have a name?

Thank you for addressing our comments and clarifying your question! Any square matrix which satisfies your equation $$A^n = \begin{cases} I, & \text{if $n$ is even} \\ A, & \text{if $n$ is odd} \end{cases}$$ is its own inverse, i.e. it is an involutory matrix. This is because if $n = 2$ you have $AA = I$, which by definition means that $A$ is its own inverse. From that simple statement you can extrapolate your formula, because if $n$ is even then $$A^{n} = A^{2m} = (AA)^{m} = I^{m} = I$$ for some integer $m$, and if $n$ is odd then $$A^{n} = A^{2k+1}=A^{2k}A=(AA)^{k}A = I^{k}A=IA=A$$ for some integer $k$. In fact, for any $2\times 2$ matrix $$\begin{pmatrix} a & b\\c & -a\end{pmatrix},$$ of which your matrix is an example, such a matrix will be involutory if $a^{2} + bc = 1$. We can verify this for your matrix $A$: $$x^{2} - (x-1)(x+1) = 1.$$

You simply have $A^2=I$, from which it follows that $A^3=AA^2=AI=A$, $A^4= AA^3=AA=A^2=I$ and so on. Of course $A^2=I$ is the same as saying that $A=A^{-1}$ so that $A$ is its own inverse.

This is sometimes stated as $A$ is self-inverse, or that $A$ is an involution, in more highfalutin language.