Unitarily equivalent $C^*$-algebra representations the situation i want to talk about is the following:
$(H_1,\varphi_1),(H_2,\varphi_2)$ irreducible representation of a $C^*$-algebra $A$. A bounded operator $T:H_1\rightarrow H_2$ such that $T\varphi_1(a)=\varphi_2(a)T$ for all $a\in A$
I am asked to prove the following implication: If $\varphi_1$ and $\varphi_2$ are NOT unitarily equivalent, then $T=0$

My ideas for so far: Proof by contradiction: Suppose $T\neq 0$, then $T^*$ exists and is bounded. I want to prove that $T$ is unitary because then we have the situation that both representations are equivalent (unitarily). But therefore i have to show that $T$ is surjective and that $(Tx,Ty)=(x,y)$ for all $x,y$, but this seems to be very difficult or is there something which i don't see and which makes the proof easier? I only know more that non-zero vectors are cyclic vectors since the representations are irreducible but i have no idea how to use it. Or: is the direct implication the better choise?

You have
$$
T^*T\varphi_1(a)=T^*\varphi_2(A)T=\varphi_1(a)T^*T.
$$
(for the second equality, note that $\varphi_1(a)T^*=T^*\varphi_2(a)$ by taking adjoints on your original equality). 
So $T^*T$ commutes with $\varphi_1(a)$ for all $a\in A$. As $\varphi_1$ is irreducible, $T^*T$ commutes with every operator in $B(H_1)$, so it is a scalar: $T^*T=\lambda I$ for some $\lambda\in\mathbb C$. If $T\ne0$, then $\lambda\ne0$. 
A similar argument shows that $TT^*$ is scalar, and since the nonzero elements in the spectra of $T^*T$ and $TT^*$ agree, we get that $TT^*=\lambda I$. 
From the positivity of $T^*T$ we get that $\lambda>0$. Then $V=T/\sqrt\lambda$ is a unitary that conjugates $\varphi_1$ and $\varphi_2$. 
