Let $V$ be a finite-dimensional complex vector space, and $U$ be any unitary that acts upon it. We want to investigate interesting subspaces of $V \otimes V$ which are invariant under all unitaries of form $I \otimes U$.
Of course, $\{0\}$ and $V \otimes V$ are invariant subspaces as normal, as are all subspaces of form $v \otimes V$ and sums thereof. Aside from these trivial cases, do any other invariant subspaces exist? If so, what are they, and if not, why?
$\DeclareMathOperator\U{U}\DeclareMathOperator\Hom{Hom}$Write $d=\dim V$. Then $V$ is a $\U(V)$-representation. In the setting of the question, the space $V\otimes V$ is also a $\U(V)$-representation, where $U\in\U(V)$ acts via $I\otimes U$. This is the same as the representation $V^{\oplus d}$, and you are asking what subrepresentations does this space have. Since $V$ is an irreducible $\U(V)$-representation, every irreducible subrepresentation will be of the form $v\otimes V$ for some $0\neq v\in V$. The reason for this is the following: if $W$ is an irreducible representation that is not isomorphic to $V$, then $\Hom(W,V^{\oplus d})=0$ by Schur's Lemma. There is no irreducible subrepresentation here to consider. On the other hand $\Hom(V,V^{\oplus d})\cong \Hom(V,V)^{\oplus d} \cong\mathbb{C}^d\cong V$. By following the isomorphisms we see that the homomorphism $V\to V^{\oplus d}$ that corresponds to $v\in V$ is exactly $v'\mapsto v\otimes v'$.
All other subrepresentations will be sums of such irreducible representations.
We first observe that every linear operator can be written as a linear combination of unitary operators. If $H$ is Hermitian, then $H=\lim_{t\rightarrow 0}(e^{iHt}-I)\cdot t^{-1}$, so $H$ is arbitrarily close to a linear combination of unitary operators. Therefore, since we are only working in a finite dimensional space, $H$ is a linear combination of unitary operators. On the other hand, since every operator can be written as $H_1+iH_2$ where $H_1,H_2$ are Hermitian, every linear operator is a linear combination of unitary operators.
For this post, suppose that $V,W$ are finite dimensional vector spaces over the same field. Then we shall say that a subspace of $V\otimes W$ is irreducible if it is irreducible with respect to all mappings of the form $(1_V\otimes A)$ where $A\in L(W)$.
Observe that the smallest irreducible subspace containing $u\in V\otimes W$ is the space consisting of all vectors of the form $(1\otimes A)u$ for some linear transformation $A\in L(W)$, and all irreducible subspaces of $V\otimes W$
Proposition: Suppose that $V,W_1,W_2$ are finite dimensional vector spaces over a field $K$. Let $(e_1,\dots,e_r)$ be a basis for $W_1$, and let $(f_1,\dots,f_s)$ be a basis for $W_2$. Suppose that $u\in V\otimes W_1,v\in V\otimes W_2$. Then we can write $u=\sum_{k=1}^ru_k\otimes e_k,v=\sum_{k=1}^sv_k\otimes f_k$. Then there is an $A\in L(W_1,W_2)$ with $(1_V\otimes A)u=v$ if and only if $\langle v_1,\dots,v_s\rangle\subseteq\langle u_1,\dots,u_r\rangle.$
Proof:
$\leftarrow$ Suppose that $\langle v_1,\dots,v_s\rangle=\langle u_1,\dots,u_r\rangle$. Then whenever $1\leq j\leq s$, we can write $v_j$ as $v_j=a_{j,1}u_1+\dots+a_{j,r}u_r$ for some constants $a_{j,1},\dots,a_{j,r}$. Therefore, let $A$ be the linear transformation where $f_j=a_{j,1}e_1+\dots+a_{j,r}e_r$ for $1\leq j\leq s$. Then let $A$ be the linear transformation where $Ae_k=\sum_{j=1}^sa_{j,k}f_j$. Then $$(1\otimes A)u=\sum_{k=1}^ru_k\otimes Ae_k=\sum_{k=1}^ru_k\otimes\sum_{j=1}^sa_{j,k}f_j=\sum_{j=1}^s\sum_{k=1}^ra_{j,k}u_k\otimes f_j=\sum_{j=1}^sv_j\otimes f_j=v.$$
$\rightarrow$ Suppose that $(1_V\otimes A)u=v$. Then let $Ae_k=\sum_{j=1}^sa_{j,k}f_j$ for $1\leq k\leq r$. Then $$v=(1_V\otimes A)u=(1_V\otimes A)\sum_{k=1}^ru_k\otimes e_k=\sum_{k=1}^ru_k\otimes Ae_k$$ $$=\sum_{k=1}^ru_k\otimes\sum_{j=1}^sa_{j,k}f_j=\sum_{j=1}^s\sum_{k=1}^ra_{j,k}u_k\otimes f_j.$$ Therefore, $v_j=\sum_{k=1}^ra_{j,k}u_k$ for $1\leq j\leq s$.
Q.E.D.
By the above result, the irreducible subspace generated by $u$ is $\langle u_1,\dots,u_r\otimes W$. As a consequence, all irreducible subspaces of $V\otimes W$ are of the form $V_1\otimes W$ where $V_1$ is a subspace of $V$.
Some comments about the partial trace.
In the case that $V,W$ are finite dimensional complex inner product spaces, we can describe the irreducible subspace generated by $u\in V\otimes W$ in terms of the partial trace $\text{Tr}_W$.
Observation: Let $V,W$ be finite dimensional complex inner product spaces where W has basis $(e_1,\dots,e_r).$ Suppose that $u\in V\otimes W$, and $u=\sum_{k=1}^ru_k\otimes e_k$. Then
$\text{Tr}_W(uu^*)=\sum_{j=1}^ru_ju_j^*$, and
$\text{Im}(\text{Tr}_W(uu^*))=\langle u_1,\dots,u_r\rangle$.
The above observation allows us to prove the following fact.
Proposition: Let $V,W$ be finite dimensional complex inner product spaces. Suppose that $u,v\in V\otimes W$. Then the following are equivalent.
$v\in \text{Im}(\text{Tr}_W(uu^*))\otimes W$
$v=(1\otimes A)u$ for some $A\in L(W)$
$\text{Im}(\text{Tr}_W(vv^*))\subseteq \text{Im}(\text{Tr}_W(uu^*))$.
Therefore, $\text{Im}(\text{Tr}_W(uu^*))\otimes W$ is the smallest invariant subspace of $V\otimes W$ containing the element $u$. This fact is related to the following standard result from quantum information theory.
Theorem: (Unitary equivalence of purifications) Suppose that $V,W$ are finite dimensional complex inner product spaces. Suppose that $u,v\in V\otimes W$. Then there is a unitary $U\in L(W)$ with $(1_V\otimes U)u=v$ if and only if $\text{Tr}_W(uu^*)=\text{Tr}_W(vv^*)$.
