Hmm...this is what I get:

$(aA + (1-a)I)^T(aA + (1-a)I) = I$

$(aA^T + (1-a)I)(aA + (1-a)I) = I$

$a^2A^TA + (a - a^2)(A^T + A) + (1-a)^2I = I$

$(a - a^2)(A^T + A) = I - a^2I - I + 2aI - a^2I = 2(a - a^2)I$.

Since $0 < a < 1$, we have $a - a^2 \neq 0$, so that:

$A^T + A = 2I$.

What does this say about what the diagonal elements of $A$ must be?

Finally, if $A_j$ is the $j$-th column of $A$, show that $\|A_j\| = 1$ (since $A$ is orthogonal). What does this force the off-diagonal elements to be?