Short version
Fix $d,t\geq 1$. Denote by $U(d)$ the set of unitary operators on $\mathbb{C}^d$. We can see $(\rho_t,\mathbb{C}^{\otimes 2t})$ is a representation of $U(d)$ where $\rho_t(U):=U\otimes\bar{U}\otimes\dotsb\otimes U\otimes\bar{U}=(U\otimes\bar{U})^{\otimes t}$ for any $U\in U(d)$. What is the irreducible decomposition of this representation?
Long version (=short version but with some ideas)
Context:
Fix $d>1$ and set $H=\mathbb{C}^d$. Denote the set of operators on $H$ by $L(d)$ and the set of unitary operators on $H$ by $U(d)\subset L(d)$. Define the unit vector $$|\psi\rangle:=\frac{1}{\sqrt{d}}\sum_i |ii\rangle\in H\otimes H,$$ or with linear algebra notation $\psi:=\frac{1}{\sqrt{d}}\sum_i e_i\otimes e_i\in H\otimes H$, where $(|i\rangle)_i=(e_i)_i$ is the canonical basis of $H=\mathbb{C}^d$.
For each, $U\in U(d)$, set $\rho(U):=U\otimes\bar{U}$. This defines a representation $(\rho,H\otimes H)$ of $U(d)$. Since $U\otimes\bar{U}|\psi\rangle=|\psi\rangle$ for any $U\in U(d)$, $G_0$ and $G_1$ are two sub-representations of $(\rho,H\otimes H)$, where \begin{equation} \begin{split} G_0&:=\mathrm{Span}(\psi),\\ G_1&:=\psi^\perp, \end{split} \end{equation} so that $H\otimes H=G_0\oplus G_1$.
The sub-representation $G_0$ is irreducible since it is $1$-dimensional. We can also prove that $G_1$ is irreducible. Here is one proof I come up with: for any $|\phi\rangle\in H\otimes H$, there exists a unique $X\in L(d)$ such that $|\phi\rangle=(X\otimes I_d)|\psi\rangle$ and therefore $$\rho(U)|\phi\rangle=(UX\otimes\bar{U})|\psi\rangle=(UXU^*\otimes I_d)|\psi\rangle.$$ We also notice if $|\phi\rangle=(X\otimes I_d)|\psi\rangle$, then $\phi\perp\psi$ is equivalent to $\mathrm{tr}(X)=0$. Therefore, an invariant subspace of $G_1$ corresponds to a subspace $V\subset\{X\in L(d): \mathrm{tr}(X)=0\}$ such that $UVU^*\subset V$, and we can prove either $V=\{X\in L(d): \mathrm{tr}(X)=0\}$ or $V=\{0\}$.
In conclusion, $H\otimes H=G_0\oplus G_1$ is an irreducible decomposition of the representation $(\rho,H\otimes H)$.
My question:
Fix $t>1$. For each $U\in U(d)$, set $\rho_t(U):=(\rho(U))^{\otimes t}=(U\otimes\bar{U})^{\otimes t}$. This again defines a representation $(\rho_t,H^{\otimes 2t})$ of $U(d)$. For any $\mathbf{b}\in\{0,1\}^t$, define $$G_{\mathbf{b}}:=G_{b_1}\otimes\dotsb\otimes G_{b_t}\subset H^{\otimes 2t},$$ where $\mathbf{b}=(b_1,\dotsc,b_t)$. We see that $G_{\mathbf{b}}$ is a sub-representation of $(\rho_t,H^{\otimes 2t})$ for evey $\mathbf{b}$, and is an irreducible one if $|\mathbf{b}|=\#\{i:b_i=1\}$ equals $1$. I do not know if $G_{\mathbf{b}}$ is irreducible for $|\mathbf{b}|>1$, and I doubt it is not.
So my question is:
What is the irreducible decomposition of $H^{\otimes 2t}$? Knowing that we have already $H^{\otimes 2t}=\oplus_{\mathbf{b}}G_{\mathbf{b}}$, which I doubt is not irreducible.