Find the normalized state and function such that the equation holds for arbitrary unitary matrix I have been recently puzzled with a problem I do not know how to solve. Here is the setting and some of my thoughts on the problem.
Given:
Let us denote the set of all unitary $d \times d$ matrices as $\mathbb{V}_d$.
$U^*$ denotes the complex conjugate of $U$, note that this does not involve transpose.
The question is to find a normalized state $|\psi\rangle$ and a function $f: \mathbb{V}_d \to \mathbb{C}$ which is a mapping from $\mathbb{V}_d$ to $\mathbb{C}$, such that the following equation:
$$ U \otimes U^* |\psi\rangle = f(U) | \psi \rangle $$ 
holds for arbitrary $U \in \mathbb{V}_d$.
Thoughts:
Assume $|\psi\rangle$ is bipartite and can be represented as $|\phi\rangle \otimes |\phi\rangle$, assume also that there exists $g:\mathbb{V}_d \to \mathbb{C}, f = g^2, g(U) = g(U^*), \forall U \in \mathbb{V}_d$.
Then, $U \otimes U^* |\psi\rangle = (U \otimes U^*) |\phi\rangle \otimes |\phi\rangle = g(U)|\phi\rangle \otimes g(U)|\phi\rangle$, which simplifies the problem to $U | \phi \rangle = g(U) |\phi\rangle$, however I am still having hard times finding such $| \phi \rangle$ and $g$. Any help or comments appreciated.
 A: Let us first decompose $ | \psi \rangle $ in the tensor product of standard bases: $ | \psi \rangle = \sum_{j,l} \Psi_{j,l} | e_j \rangle \otimes | e_l \rangle$. 
Then, 
$$\begin{aligned} U \otimes U^* | \psi \rangle &=
 \sum_{j,l} \Psi_{j,l} U | e_j \rangle \otimes U^* | e_l \rangle \\ 
&= \sum_{i,k} \sum_{j,l} \Psi_{j,l} U_{i,j} | e_j \rangle \otimes U_{k,l}^*| e_l \rangle\\ 
&= \sum_{i,j,k,l} \Psi_{j,l} U_{i,j} U_{k,l}^* | e_j \rangle \otimes | e_l \rangle\\ 
&= \sum_{i,j,k,l} U_{i,j} \Psi_{j,l} U_{l,k}^{*^T} | e_j \rangle \otimes | e_l \rangle\\ 
&=\sum_{i,j,k,l} U_{i,j} \Psi_{j,l} U_{l,k}^{\dagger} | e_j \rangle \otimes | e_l \rangle.\end{aligned}$$
The problem asks the last line being equal to $f(U) | \psi \rangle$ which can be similarly decomposed as
$$ \sum_{j,l} f(U) \Psi_{j,l} | e_j \rangle \otimes | e_l \rangle , $$
which yelds
$$ U \Psi U^{\dagger} = f(U) \Psi . $$
Now let's use the following properties of unitary matrices:
$$ U^{\dagger} = U^{-1} , $$
$$ |\text{det}(U)| = 1 , $$
and
$$ \text{det}(AB) = \text{det}(A)\text{det}(B) . $$
Now we have
$$ |\text{det}(U \Psi U^{\dagger})| = |\text{det}\Psi| = |f(U)\text{det}\Psi| , $$
which lets us set $f(U)=1$.
The only thing left to deal with is finding a matrix $\Psi$ such that 
$$ U \Psi U^{\dagger} = \Psi , $$
which can be equivalently rewritten as
$$ U \Psi = \Psi U . $$
Looking at definitions of Unitary Group $U(n)$ and Center of a Group on Wikipedia one can convince him/herself that $\Psi$ is the Center of $U(d^2)$ and thus can be written as $e^{i\phi}I$ which implies the state to be
$$ | \psi \rangle = e^{i \phi} \sum_{i} | e_i \rangle \otimes | e_i \rangle .$$
The only thing left is to normalize our state by $\sqrt{d}$.
