First, if you take the fundamental representation (representation $N$) of $SU(N)$ made of $N$ objects $\varphi^i$, the transformation law is :

$\varphi^i \to U^i{}_j \varphi^j$.

By taking the complex conjugate, you get : $\varphi^{*i} \to (U^*)^i{}_j \varphi^{*j}= (U^\dagger)^j{}_i \varphi^{*j}$.

Now, looking at the last expression with $U^\dagger$, one sees that it is more practical to define objects $\varphi_i$, wich transform like $\varphi^{*i}$ :

$\varphi_{i} \to (U^\dagger)^j{}_i \varphi_{j}$,

This is the representation $\bar N$

Now clearly, when you make the product of the two representations $N$ and $\bar N$, you have a representation $T^i_j$ which transforms as $\varphi^i\varphi_j$ :

$T^i_j \to (U)^i_k (U^\dagger)^l_j T^k_l$

Secondly, you cannot symmetrize or anti-symmetrize the representation $N \otimes \bar N$, that is $T^i_j$, because the indices $i$ and $j$ have a different nature, and correspond to different representations.

Now, if you consider the representation $N \otimes N$, that is some representation $S^{ij}$, then here you may separe in a symmetric and anti-symmetric part, for instance, you have :

$3 \otimes 3 = 6 \oplus \bar 3 $

The $6$ is the symmetric part, while the $\bar 3$ is dual (equivalent) to the anti-symmetric part, thanks to the Levi-Civita tensor : $\varphi_i = \epsilon_{ijk} \varphi^{jk}$

First, if you take the fundamental representation (representation $N$) of $SU(N)$ made of $N$ objects $\varphi^i$, the transformation law is :

$\varphi^i \to U^i{}_j \varphi^j$.

By taking the complex conjugate, you get : $\varphi^{*i} \to (U^*)^i{}_j \varphi^{*j}= (U^\dagger)^j{}_i \varphi^{*j}$.

Now, looking at the last expression with $U^\dagger$, one sees that it is more practical to define objects $\varphi_i$, wich transform like $\varphi^{*i}$ :

$\varphi_{i} \to (U^\dagger)^j{}_i \varphi_{j}$,

This is the representation $\bar N$

Now clearly, when you make the product of the two representations $N$ and $\bar N$, you have a representation $T^i_j$ which transforms as $\varphi^i\varphi_j$ :

$T^i_j \to (U)^i_k (U^\dagger)^l_j T^k_l$

Secondly, you cannot symmetrize or anti-symmetrize the representation $N \otimes \bar N$, that is $T^i_j$, because the indices $i$ and $j$ have a different nature, and correspond to different representations.

Now, if you consider the representation $N \otimes N$, that is some representation $S^{ij}$, then here you may separe in a symmetric and anti-symmetric part, for instance, you have :

$3 \otimes 3 = 6 \oplus \bar 3 $

The $6$ is the symmetric part, while the $\bar 3$ is dual (equivalent) to the anti-symmetric part, thanks to the Levi-Civita tensor : $\varphi_i = \epsilon_{ijk} \varphi^{jk}$

[EDIT]

Due to OP comments, some precisions :

You have $U^\dagger = (U^*)^T$, where $T$ means transposed operation. Transposition means exchange of the row and columns of the matrix, that is exchange of the $i$ and $j$ indices. If you put the row indice as an upper indice and the column indice as a lower indice, then the exchange necessarily will put the row indice as a lower indice, and the column indice as a upper indice. Your notation $(U^*)^i{}_j = (U^\dagger)_j{}^i$ is a not-too-good equivalent notation, I say not-too-good, because you loose the orginal meaning that I describe above .About the representations, this is a different thing (these are not the same $i$ and $j$...), the upper indice transforms as a $N$ representation, and the lower indice transforms as a $\bar N$ representation, so it is like apples and bananas, you can only symmetrize or anti-symmetrize equivalent quantities which transform in the same manner ($2$ apples or $2$ bananas), but not $1$ banana + $1$ apple