linear algebra – dimension of a unitary matrix

Suppose I have a hilbert space $ V $ of dimension $ N = $ 7 which carries a unitary representation $ rho $ of $ C_ {2v} $ group of points.

In particular, suppose $ rho $ divides the space $ V $ so that $ V cong 2A + B_1 + 2B_2 + 2B_3 $.

The variety of unit operators $ U $ sure $ V $ commute with $ rho $ should have a dimension $ 2 ^ 2 + 2 ^ 2 + 2 ^ 2 + 1 ^ 2 = $ 13 since such $ U $ can be broken in $ U_A + U_ {B_1} + U_ {B_2} + U_ {B_3} $ where each component acts only on the corresponding subspace.

What is the size of the variety of "superoperators" $ tilde {U} $ formed of the $ U $ by the rule
$$
X to tilde {U} X: = U X U ^ Dagger + U & # 39; {} ^ Dagger
$$

Or $ U # stag { pi} {2}} U ^ dagger _ { frac { pi} {2}} $ and $ O _ { frac { pi} {2}} $ is the element of a representation $ rho & # 39; $ of $ C_ {4v} $ group of points corresponding to $ 90 ^ circ $ rotation.

$ rho $ is obtained from $ rho & # 39; $ by restriction of $ C_ {4v} $ to the subgroup $ C_ {2v} $.

I understand that the dimension can not be bigger than $ 13 but it's difficult for me to know if the operation of "quadrature" $ U to U ( cdot) U ^ Dagger $ or the $ 90 ^ circ $ The "averaging" serves to reduce the dimensionality of the space of possibilities $ tilde {U} $.

Also what is the dimension of the space of $ tilde {U} $ if we only allow it to operate on the operator space totally symmetrical with respect to $ rho $ (that is to say, symmetry $ A $)?
I believe this is equivalent to defining the $ tilde {U} $ to be instead
$$
X to tilde {U} X: = sum_ {n = A, B_1, B_2, B_3} U_n X U_n ^ dagger + U & nbsp; _n X U _n {} ^ dagger
$$