Show that $Vert vVert^2= langle v,AA' vrangle iff AA'=I$

Let's prove $|Av| = |v|$ for all $v in V$ iff $langle Av, Awrangle = langle v, w rangle$ for all $v, w in V$. This plus the hints should be enough for you to finish it.

If $langle Av, Aw rangle = langle v, w rangle $for all $v, w in V$, then set $v = w$ to get $|Av|^2 = langle Av, Av rangle = langle v, v rangle = |v|^2$, so $|Av| = |v|.$

If $|Av| = |v|$ for all $v in V$, then $|A(v+w)| = |v+w|$ for all $v, w in V$. Notice

$$ |A(v+w)|^2= langle A(v+w), A(v+w) rangle = langle Av + Aw, Av + Aw rangle = langle Av, Avrangle + 2langle Av, Aw rangle + langle Aw, Aw rangle = |v|^2 + 2 langle Av, Aw rangle + |w|^2,$$ $$ |v+w|^2 = langle v+w, v+w rangle = |v|^2 + 2langle v, w rangle + |w|^2.$$ Setting these equal to each other and doing some algebra, we have

$$ 2 langle Av, Aw rangle = 2 langle v, w rangle implies langle Av, Aw rangle = langle v, w rangle.$$

Edit: I'll also solve your problem in the comments. Fix $y$ and $z$. We claim $langle x, y rangle = langle x, z rangle$ for all $x$ iff $y = z$. The hard part is the implication, which we do now. If $langle x, y rangle = langle x, z rangle$, then $langle x, y- z rangle = 0$ for all $x$. In particular, $langle y-z, y-z rangle = |y-z|^2 = 0$. But recall $|y-z|^2 = 0$ if and only if $y - z =0$, or $y = z$.