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what can we say about $\det(AA^t)$ when $(AA^t)^r=I$, where $A$ is $m\times n$ real matrices, $r$ is a natural number.

If $B=AA^t$ then $(\det(B))^r=\det I=1\Rightarrow \det B=\pm 1$ if $r$ is even and $ 1$ if $r$ is odd. am I right?

Myshkin
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For a matrix $X$, $\mathrm{det}(X^n) = \mathrm{det}(X)^n$, so here let $X=AA^t$ (I suppose that $A^t$ stands for the transpose of $A$ and $r>0$) and

$\mathrm{det}(X^r)=1$

therefore,

$\mathrm{det}(X)^r = 1$.

But since $X=AA^t$ is positive semidefinite, its determinant is nonnegative, so we keep only the positive root, that is

$\mathrm{det}(X)=1$

Pantelis Sopasakis
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    And since $X$ (aka $B$) is diagonalizable, and all its eigenvalues are equal to $1$, it can only be the identity matrix. – Erick Wong Jun 23 '17 at 16:53