The Hilbert A-modules $A$ and $A\oplus A$ are unitarily equivalent iff $\mathcal{O}_2\subseteq M(A)$ The following exercise appears in "Hilbert $C^*$-modules" book by E. Christopher Lance:
Definition: $E$ and $F$ are unitarily equivalent Hilbert $A$-modules if there exists a unitary element $u\in L(E,F)$, where $L(E,F)$ denotes the set of $A$-linear maps that are adjointable.
Exercise: Show that $A$ and $A\oplus A$ are unitarily equivalent Hilbert $A$-modules if and only if $M(A)$ (the multiplier algebra) contains a unital $C^*$-subalgebra isomorphic to the Cuntz algebra $\mathcal{O}_2$.
This should be a basic exercise, but I have no idea how to approach it.
Thank you for any help!
 A: If you have $\mathcal O_2\subset M(A)$, you can take two orthogonal isometries $V,W\in M(A)$ with $V^*V=W^*W$, and $VV^*+WW^*=I$. Now you can define $$u:A\to A\oplus A$$ by 
$$ua= V^*a \oplus W^*a .$$ This is well-defined because $V,W\in M(A)$. We need to check that $u$ is adjointable:
$$
\langle ua,b\oplus c\rangle_{A\oplus A}=\langle V^*a \oplus W^*a ,b\oplus c\rangle_{A\oplus A}
=b^*V^*a +c^*W^*a =\langle a,Vb+Wc\rangle_A.
$$
So $u^*:A\oplus A\to A$ is given by $u^*(b\oplus c)=Vb+Wc$. In particular, 
$$
u^*u\,a=VV^*a+WW^*a=a,
$$
so $u^*u=I$. And \begin{align} uu^*(b\oplus c)&=u(Vb+Wc)=(V^*Vb+V^*Wc)\oplus (W^*Vb+W^*Wc)\\
&=V^*Vb\oplus W^*W c=b\oplus c.\end{align} So $uu^*=I$. Thus $u\in L(A,A\oplus A)$ is a unitary. 

Conversely, suppose that $u:A\to A\oplus A$ is a unitary. Note that $M(A)=L(A,A)$. Define $V,W\in M(A)$ by $$ Va=u^*(a\oplus 0),\ \ \ Wa=u^*(0\oplus a). $$ Then
$$
\langle V^*Va,b\rangle_A=\langle Va,Vb\rangle _A=\langle u^*(a\oplus 0),u^*(b\oplus 0)\rangle _A
=\langle (a\oplus 0),(b\oplus 0)\rangle _{A\oplus A}=\langle a,b\rangle. 
$$
It follows that $V^*V=I$, and similarly $W^*W=I$. Also,
$$
\langle W^*Va,b\rangle_A=\langle Va,Wb\rangle _A=\langle u^*(a\oplus 0),u^*(0\oplus b)\rangle _A
=\langle (a\oplus 0),(0\oplus b)\rangle _{A\oplus A}=0+0=0.
$$
We have 
$$
\langle V^*a,b\rangle_A=\langle a,Vb\rangle_A=\langle ua,(b,0)\rangle_{A\oplus A}=\langle \pi_1ua,b\rangle_A,
$$
so $V^*a=\pi_1ua$, where $\pi_1$ is the projection onto the first coordinate. Similarly, $W^*a=\pi_2ua$. Then
\begin{align}
(VV^*+WW^*)a=V\pi_1ua+W\pi_2ua=u^*(\pi_1ua,0)+u^*(0,\pi_2ua)=u^*ua=a,
\end{align}
and so $VV^*+WW^*=I$. 
Thus $\mathcal O_2\simeq C^*(V,W)\subset M(A)$.
A: I think that I have an idea for the converse direction (without the assumption that $A$ is unital):
Let $u: A\oplus A\to A$ be the given unitary element in $L(A\oplus A,A)$.
Define $p_1:A\to A\oplus A$ to be $p_1(a)=(a,0)$, then it is easy to check that $p_1\in L(A,A\oplus A)$ and $p_1^* \in L(A\oplus A,A)$ is the natural projection $p_1^*(a_1,a_2)=a_1$. 
Similarly define $p_2\in L(A,A\oplus A)$, and observe that $p_1^*p_1=p_2^*p_2=I_{L(A)}$
Then we have $u\circ p_1\in L(A)$ and $u\circ p_2\in L(A)$.
As Martin mentioned in the comment above, we know that $M(A)=L(A)$.
Now, $(u\circ p_1)^*(u\circ p_1)=p_1^*u^*up_1=p_1^*p_1=I_{L(A)}$ and similarly $(u\circ p_2)^*(u\circ p_2)=I_{L(A)}$ .
Moreover, $(u\circ p_1)(u\circ p_1)^*+(u\circ p_2)(u\circ p_2)^*=u(p_1p_1^*+p_2p_2^*)u^*=uu^*=I_{L(A)}$.
By the universal property of $\mathcal{O_2}$ and simplicity, we have $C^*(u\circ p_1, u\circ p_2)$ is an isomorphic $C^*$-subalgebra of $M(A)$.
