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Let $(E,langlecdot;,;cdotrangle)$ be a complex Hilbert space and $mathcal{L}(E)$ be the algebra of all operators on $E$.

Assume that $Tin mathcal{L}(E)$ and satisfy the following property (P):

$$langle x,yrangle=0Longrightarrow langle Tx,Tyrangle=0,$$
for all $x,yin E$.

I want to prove under the property $(P)$ that there exists $cgeq0$ such that
$$|Tx|=c|x|,$$
for all $xin E$.

Note that, if there exist $x, yin Esetminus{0}$ such that $|Tx|=alpha|x|$ and $|Ty|=beta|y|$
for some $0leqalpha <beta$, it can be seen that $langle x,yrangleneq0$.

The claim as stated is false, because the operator $T=0$ provides a counterexample.

I assume for the rest that $Tneq 0$.

You wrote that

Note that, if there exist $x, yin Esetminus{0}$ such that $|Tx|=alpha|x|$ and $|Ty|=beta|y|$
for some $0<alpha< beta$, it can be seen that $langle x,yrangleneq0$.

This is already a very good start.
However, this statement is also true if you allow $alpha=0$
(although i do not know your proof of your statement, I would guess it could be modified to allow $alpha=0$).
Thus we can assume that $ker T={0}$.

Let $xin Esetminus{0}$ be an arbitrary point and let $c>0$ be such that $|Tx|=c|x|$.
For an element $zin Esetminus{0}$ with $langle x,zrangle=0$,
we can use your observation to obtain that $|Tz|=c|z|$.
(Otherwise we would have have $|Tz|=alpha|z|$ for some $alphaneq c$.
Then your observation would imply $langle x,zrangle neq0$.)

Let $yin E$ be given.
Then we can find a decomposition $y=alpha x+z$ for $alphainmathbb C$ and $zin {x}^perp$.
Note that it follows that $alpha Txperp Tz$.
We calculate
$$
|Ty|^2=|alpha Tx+Tz|^2 = |alpha|^2|Tx|^2+|Tz|^2+ langle alpha Tx,Tzrangle
+ langle Tz,alpha Txrangle
\= |alpha|^2 c^2 |x|^2+c^2|z|^2 = c^2|alpha x+z|^2 = c^2|y|^2.
$$

Thus it holds for all $yin E$.

Suppose there exists $x,yneq 0$ with $langle x,yrangle =0$ and $|T(x)|=a|x|, |T(y)|=b|y|, aneq b$. We can assume $|x|=|y|=1$.

$$langle x+y,x-yrangle=0, langle T(x+y),T(x-y)rangle$$ $$ =langle T(x),T(x)rangle-langle T(x),T(y)rangle+langle T(y),T(x)rangle-langle T(y),T(y)rangle=a^2-b^2neq 0$$ contradiction.

Let $xneq 0$, $yin Vec(x)^{perp}$, $|T(x)|=c|x|, |T(y)|=c|y|$. For every $zin H$, $z=ux+vy, yin Vect(x)^{perp}$, implies $|T(z)|^2=|T(ux+vy)|^2=langle T(ux)+T(vy),T(ux)+T(vy)rangle=c^2|z|^2$.

Assuming the statement you made at the end we can complete the proof as follows. We first show that $T$ is one-to-one unless it is the zero operator. Suppose $Tx =0, xneq 0$. Let $y neq0$. If $langle x,yrangle neq 0$ and $a= frac {langle x,yrangle } {|y|^{2}}$then $x-ay$ and $y$ are orthogonal. The hypothesis now tells us that $Ty=0$. On the other hand if $langle x,yrangle = 0$ then there exist $aneq 0, bneq 0$ such that $x-ay$ and $x-by$ are orthogonal which again gives $Ty=0$. Hence $T$ is the zero operator in which case we can take $c=0$. Thus $T neq 0$ implies $T$ is one-to -one. Let ${e_i}$ be an orthonormal basis (not necessarily countable). It follows that $|Te_i|$ is independent of $i$ (unless $T(e_i)=0$). Also $T(e_i)$ are orthogonal. This immediately gives the result with $c=|Te_i|$. Just expand any element in terms of the basis and use orthogonality of $T(e_i)$'s. [Proof of the fact that $ |Te_i |neq |Te_j |$ cannot occur: suppose $|Te_i |< |Te_j |$. Take $x =e_i,y=e_j, alpha =|Te_i |, beta =|Te_j |$. We get the contradiction that $langle e_i,e_jrangle neq 0$].