This is really a 3 part question:-
Are there infinite involutory matrices of a given order $n\times n$?
This article claims that there are infinite involutory matrices, but I think it only claims that only by counting identity matrices for all possible orders.
If yes, then does this include both real and complex matrices? or only one of these 2 types of matrix constitutes and infinite number of involutory matrices?
The wikipedia article mentions a few matrices that are not $I_{n}$ or $(-I_{n})$ of both complex or real-only element types but does not comment on the number of possible matrices.
If not, is it possible to calculate number of involutory matrices for the order $n\times n$?
I am a high school student so if you can give a simpler proof would be appreciated. (However please do tell even if it is not possible using elementary mathematics.)
Conceptually, consider the linear transformations defined by rotating the space through an angle $\theta$, flipping one of the coordinate axes, and rotating back. This is clearly an involution, and since it is a linear transformation, there is a matrix representing it. They are also all distinct, because for each $\theta$, there is a different unit vector that gets flipped around exactly (it's the one that gets rotated to the flipped axis in the first step). Since there are infinitely many angles $\theta$, there are infinitely many such transforms.
For a more general version of this argument, let $D$ be a diagonal matrix whose nonzero elements are all $\pm 1$ that is not $I$ or $-I$ and let $O$ be an orthogonal matrix. Then $$ T = O^TDO $$ is an involution, and this involution will in general be distinct for distinct $O$.