A composite system AB, its initial state is a product state of A and B given by 
$(I_A\otimes \rho_B)$ (A in a completely mixed state). The composite system undergoes a unitary operation $U_{AB}$. My calculation seems to show that we have

$$\mathrm{Tr}_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})=I_A$$

since if the dimensions of A and B are m and n respectively, then we have:

$$I_A\otimes \rho_B =  \left[ {\begin{array}{cccc}
    \rho_B & & &  \\    & \rho_B & & \\  & & \ddots&  \\ & & & \rho_B  \end{array} } \right]
,\quad
U_{AB} = \left[ {\begin{array}{cccc}
    u_{11} & u_{12} & \cdots & u_{1m}  \\ 
    u_{21} & u_{22} & \cdots & u_{2m}  \\ 
    \vdots & \vdots & \ddots & \vdots  \\ 
    u_{m1} & u_{m2} & \cdots & u_{mm}  \\ 
 \end{array} } \right]$$

where $u_{ij}$ is a $n \times n$ matrix


Then $O_{AB}=U_{AB}(I_A\otimes \rho_B)U_{AB}^{+}$ is given by

$$O_{AB} = \left[ {\begin{array}{cccc}
    o_{11} & o_{12} & \cdots & o_{1m}  \\ 
    o_{21} & o_{22} & \cdots & o_{2m}  \\ 
    \vdots & \vdots & \ddots & \vdots  \\ 
    o_{m1} & o_{m2} & \cdots & o_{mm}  \\ 
 \end{array} } \right],$$

where $o_{ij}=\sum_{k}u_{ik}\rho_B u_{jk}^{+}$, and the $(i,j)$ item of $\mathrm{Tr}_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})$ is given by $\mathrm{Tr}(o_{ij})$.

And due to the fact that $U_{AB}$ is unitary, so $\sum_{k}u_{ik}u_{jk}^{+}=\sigma_{ij}I_{n\times n}$. This leads to 

$Tr(o_{ij})=Tr(\sum_{k}u_{ik}\rho_B u_{jk}^{+})=Tr(\sum_{k}u_{jk}^{+}u_{ik}\rho_B)=Tr((\sum_{k}u_{jk}^{+}u_{ik})\rho_B)=\sigma_{ij}$

we can easily get 

$$\mathrm{Tr}_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})=I_A$$

This means that starting from a completely mixed state $I_{A}$, subsystem A will stay in a completely mixed state no matter how it's jointly evolved with another system B if A and B are in an initial product state. 

Is my calculation correct? Or I made a mistake somewhere? 

PS: Finally I found I am wrong with my calculation. Please refer to my own comments on my question.