Determinant of an orthogonal matrix

Question

The question goes like this,
For a square matrix A of order 12345, if det(A)=1 and AA'=I (A' is the transpose of A) then det(A-I)=0
(I have to prove it if it is correct and provide a counterexample if wrong).

So all that I know is that the given matrix is an orthogonal matrix. I tried to multiply det(A-I) by det(A') and det(A) from the left and right and use the properties of determinants to reduce the expression but that led me nowhere. Is there something that I am missing from the properties of orthogonal matrices or matrices in general? Any hint or reference will really be appreciated. Please

Answer 1

Note that $$\det(A-I) = \det(A - AA^\top) \stackrel{(\ast)}{=} \det(I-A^\top) = (-1)^n \det(A^\top - I) = (-1)^n \det(A-I),$$where in $(\ast)$ we use that $\det(A) = 1$. This means that if $n$ is odd, then $\det(A-I) = 0$. Now, is $12345$ odd?