Is it true a proposition for $3 \times 3$ real matrices $A$ that if for a some nonzero vector $v$ we have $Av \neq v$ and at the same time $(A^TA)v=v$ then for any vector $u$ from $\mathbb{R}^3$ we have $(A^TA)u=u$ (i.e. $A$ is orthogonal) ?
Do we need assumption about full rank of the matrix $A$ for this goal ?
Can we extend the proposition for bigger dimensions ?
On
This is wrong.
Take any orthogonal matrix $M\in \Bbb R^{2\times 2}$, and extend it into a $3\times 3$ matrix using a column of ones and a row of zeroes:
$$\left[\matrix{\left[M\right] & \matrix{1\\1}\\ \matrix{0&0}&1}\right]$$
The result is not an orthogonal matrix, but it has full rank and even determinant $1$, and most of the time $v=\left(\matrix{1\\0\\0}\right)$ will satisfy your requirements.
On
That's not true. Take any symmetric matrix $A$ for which $-1$ is an eigenvalue; then if $v$ is an eigenvector for $-1$, we have $Av=-v\neq v$ but $A^TAv=A^2v=-Av=v$. But the other two eigenvalues could be anything, so the matrix need not be orthogonal. Note that this even works with $2\times 2$ matrices.
Your claim fails even if $A$ is of full rank: for example, consider $$A=\begin{bmatrix}-1&0&0 \\ 0&2&0\\ 0&0&1\end{bmatrix}\text{ and } \mathbf{v} = \begin{bmatrix} 1\\0\\0\end{bmatrix}.$$ Then $A^TA = A^2$ and you can easily find $\mathbf{u}$ such that $A^TA\mathbf{u}\neq\mathbf{u}$.