If we have the vector $$V = \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) = \begin{pmatrix}1\\0\\0\\1\end{pmatrix}$$
where $|00\rangle = \begin{pmatrix}1 \\ 0\end{pmatrix} \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}$ and $|11\rangle = \begin{pmatrix}0 \\ 1\end{pmatrix} \otimes \begin{pmatrix}0 \\ 1\end{pmatrix}$. It can quickly be shown that
$$(A \otimes I) |V\rangle = (I \otimes A^{T}) |V\rangle$$
holds for any arbitrary 2x2 matrix A, if $A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$, and $I$ is the 2x2 identity matrix.
The resulting vector is just $\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$
my question is then: How can i use only this knowledge from the part above, to deduce that the following must be true for all unitary matrices U? $V$ is the very same vector as defined initially:
$$(U \otimes \bar{U})|V\rangle = |V\rangle$$
$\bar{U}$ denotes the complex conjugated. I have been stuck on this for a while now, even though it might be rather obvious to some. What information from the first part of my question is enough to deduce that the above equation must be true for all unitary matrices U?
I have tried to write out ($(U \otimes \bar{U})$), but this does not get me any closer:
$$\begin{pmatrix}aa^{*} & ab^{*} & ba^{*} & bb^{*} \\ ac^{*} & ad^{*} & bc^{*} & bd^{*} \\ ca^{*} & cb^{*} & da^{*} & db^{*} \\ cc^{*} & cd^{*} & dc^{*} & dd^{*}\end{pmatrix}\begin{pmatrix}1\\0\\0\\1\end{pmatrix}=\begin{pmatrix}aa^{*}+bb^{*}\\ac^{*}+bd^{*}\\ca^{*}+db^{*}\\cc^{*}+dd^{*}\end{pmatrix}$$
I can see that i get the terms $aa^{*}$ and $dd^{*}$, which is fine - but i have no arguments as to why all the other terms should vanish. I think this might be the wrong approach, and there is something much simpler i am missing. I feel i am going around in circles, coming close to an understanding, but still far from. Any hints and/or help would be appreciated!
Consider a general unitary matrix $U = \begin{pmatrix}a & b \\-e^{i\phi}\bar b & e^{i\phi}\bar a\end{pmatrix}$. Then: \begin{align} (U\otimes\bar U)(|00\rangle+|11\rangle) &= U\begin{pmatrix}1 \\ 0\end{pmatrix} \otimes\bar U\begin{pmatrix}1\\0\end{pmatrix} + U\begin{pmatrix}0\\1\end{pmatrix}\otimes \bar U\begin{pmatrix}0\\1\end{pmatrix} \\ &=\begin{pmatrix}a \\ -e^{i\phi}\bar b\end{pmatrix} \otimes \begin{pmatrix}\bar a \\ -e^{-i\phi}b\end{pmatrix} + \begin{pmatrix}b \\ e^{i\phi}\bar a\end{pmatrix} \otimes \begin{pmatrix}\bar b \\ e^{-i\phi}a\end{pmatrix} \\ &= \begin{pmatrix}a\bar a & -e^{-i\phi}ab\\-e^{i\phi}\bar b\bar a & \bar b b\end{pmatrix} + \begin{pmatrix}b\bar b & e^{-i\phi}ab\\e^{i\phi}\bar b \bar a & \bar a a\end{pmatrix} \\ & = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\\ & = \begin{pmatrix}1 \\ 0\end{pmatrix} \otimes \begin{pmatrix}1 \\ 0\end{pmatrix} + \begin{pmatrix}0 \\ 1\end{pmatrix} \otimes \begin{pmatrix}0 \\ 1\end{pmatrix} \\ & = |00\rangle + |11\rangle \end{align}
If you want to do it using your vector $\begin{pmatrix}a\bar a+b\bar b\\a\bar c+b\bar d\\c\bar a+d\bar b\\c\bar c+d\bar d\end{pmatrix}$ in the edited version of your question, note that a unitary matrix $U = \begin{pmatrix}a & b \\c & d\end{pmatrix}$ must satisfy:
$$UU^* = \begin{pmatrix}a & b \\c & d\end{pmatrix}\begin{pmatrix}\bar a & \bar c \\\bar b & \bar d\end{pmatrix} = \begin{pmatrix}a\bar a + b \bar b & a\bar c + b \bar d \\\bar a c + d\bar b & c\bar c + d \bar d\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$ So your vector must be $$(U\otimes \bar U)|V\rangle = \begin{pmatrix}1 \\ 0 \\ 0 \\ 1\end{pmatrix}$$
Using the first part of your question (which I forgot about), you get \begin{align}(U\otimes\bar U)|V\rangle &= (U\otimes I)(I\otimes \bar U)|V\rangle \\ &= (U\otimes I)(U^*\otimes I)|V\rangle \\ &= (UU^*\otimes I)|V\rangle \\ &= (I\otimes I)|V\rangle \\ &= |V\rangle\end{align}