Unitary "self equivalence" implies proportional to identity. I want to show that a linear operator $T:\mathcal{H}_A\rightarrow\mathcal{H}_B$ of (finite dimensional) Hilbert spaces is proportional to an isometry, that is $T^\dagger T=c \cdot\boldsymbol{1}_A$, if for any unitary $U$ on $\mathcal{H}_A$ there exists a unitary $U^\prime$ on $\mathcal{H}_B$ s.t. $TU=U^\prime T$.
I have seen that the supposition implies $T^\dagger T= U^\dagger \left(T^\dagger T\right) U$, so $T^\dagger T$, in a sense, is unitarily self equivalent, with an arbitrary unitary. This already implies in a handwavy way that $T^\dagger T$ should be proportional to $\boldsymbol{1}_A$, but I cannot seem to find a convincing argument.
Any pointers would be greatly appreciated.

Edit: The motivation is that this is stated in this paper on page 4, where it's called "useful", so I wanted to check it.
 A: Not the most elegant solution, but this is the first argument I came up with. Let $e_1,\ldots, e_n$ be an orthonormal base of $H_A$. First consider the unitary transformation determined by
$$
Ue_k = e^{i \theta_k} e_k
$$
where all $\theta_n$ differ from each other and $1 \leq k \leq n$. You already observed that $T^\dagger T = U^\dagger T^\dagger T U$. So if we set $T^\dagger T = A$, we find
$$
\langle A e_k, e_l \rangle = \langle A Ue_k, U e_l \rangle = e^{i (\theta_k -\theta_l)}\langle A e_k, e_l \rangle.
$$
Since $e^{i (\theta_k -\theta_l)} \neq 1$ if $k \neq l$, we must have that $\langle A e_k, e_l \rangle = 0$ if $k \neq l$. In other words $A e_k = c_k e_k$ for some constants $c_k$ with $1 \leq k \leq n$. 
Now fix some $1 < k \leq n$. Consider the unitary transformation determined by $Ue_1 = e_k$, $Ue_k = e_1$ and $U e_l = e_l$ for all $l \neq 1$ and $l \neq k$. Then we see
$$
c_k = \langle Ae_k, e_k \rangle = \langle U^\dagger A Ue_k, e_k \rangle = c_1.
$$
So we find for $c:= c_1$, that $A = c \cdot \mathbb{1}_A$. 
