I am going through my lecture nots in linear algebra and during a proof we state that
$||{Ax}||_2^2=<Ax, Ax>=<x, A^TAx>=<x,x>=||x||_2^2$ if and only if $A^TA=E$ with $E$ being the identity matrix. Unfortunately, I do not understand why from $||Ax||_2^2=||x||_2^2$ it has to follow that $A^TA$ has to be the identity matrix. Would not any isometric matrix satisfy this condition?
First of all, I assume that when you make a statement about $x$, you mean that this statement should hold for every $x \in \Bbb R^n$ (where $n$ is the size of the square matrix $A$).
Second, note that the question in your title is distinct from the question in the body of your post. It is true that if $\|Ax\| = \|x\|$ holds for all $x \in \Bbb R^n$, then $A^TA$ must be the identity matrix. It is not true that if $\|Ax\| = \|x\|$ holds for all $x \in \Bbb R^n$, then $A$ is the identity matrix.
Sticking to the correct statement, it suffices to show the following: if $\langle x,A^TA x\rangle = \langle x,x \rangle$ holds for all $x \in \Bbb R^n$, then it must hold that $A^TA = E$. To that end, first observe that $A^TA$ is symmetric. Indeed, the fact that $(PQ)^T = Q^TP^T$ holds in general means that we have $$ (A^TA)^T = A^T A^{TT} = A^TA. $$ We see that $$ \langle x, A^TAx \rangle = \langle x, x \rangle \quad \text{ for all } x\in \Bbb R^n \implies\\ \langle x, A^TAx \rangle - \langle x, Ex \rangle = 0 \quad \text{ for all } x\in \Bbb R^n\implies\\ \langle x,(A^TA - E)x \rangle = 0 \quad \text{ for all } x\in \Bbb R^n. $$ Now, Let $M = A^TA - E$, and let $m_{jk}$ denote the $j,k$ entry of $M$. Because $E$ and $A^TA$ are symmetric, $M$ is symmetric. Let $e_j \in \Bbb R^n$ denote the vector with a $1$ as its $j$th entry and $0$s elsewhere. We note that $$ 0 = \langle e_j,M e_j \rangle = m_{jj}. $$ Thus, the diagonal entries of $M$ are all equal to zero. Similarly, for $j \neq k$, $$ \begin{align} 0 &= \langle e_j + e_k, M(e_j + e_k) \rangle = m_{jj} + m_{jk} + m_{kj} + m_{kk} \\ & = 0 + m_{jk} + m_{kj} + 0 = 2 m_{jk} \end{align} $$ where $m_{kj} = m_{jk}$ holds because $M$ is symmetric. Thus, we see from the above that for all $j \neq k$, $m_{jk} = 0$. That is, the off-diagonal entries of $M$ are also equal to zero.
Thus, $M$ is the zero matrix, which is to say that $$ A^TA - E = 0 \implies A^TA = E, $$ which was what we wanted.