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I am having some trouble understanding why $\int d\psi (| \psi \rangle \langle \psi | )^{\otimes ^2}\propto \ I+$ SWAP , where $|\psi \rangle =U|\psi _0\rangle$ are Haar random states and $d\psi $ is the Haar measure.
I am somewhat convinced that $\int dU U\otimes U^\dagger \propto\ $SWAP. Not sure how to use that fact to understand the aforementioned integral. A reference or explanation would be appreciated.
I guess the gist of the question here is the relation between expressions like $\int d\psi \,\mathbb{P}_\psi^{\otimes t}$, $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$, and expressions like $\int dU \, U^{\otimes t}\otimes (\bar U)^{\otimes t} $, for some $t$. The former object is already discussed in Density matrices of multiples copies of a single Haar-Random state, and in the many links therein.
A good way to think about expressions like $\int dU \, U^{\otimes t}\otimes (\bar U)^{\otimes t} $ is that they correspond to quantum maps. More precisely, consider a quantum map of the form $$\Phi_U(X) = U^{\otimes t}X (U^\dagger)^{\otimes t},$$ for some operator $U$. Then its natural representation takes the form $$K(\Phi_U) = U^{\otimes t}\otimes (\bar U)^{\otimes t}.$$ The "natural representation" of a map $\Phi$ is here understood as the way it acts on vectorized states. That is, explicitly, an operator with elements $\langle i,j|K(\Phi)|k,\ell\rangle = \langle i|\Phi(|k\rangle\!\langle \ell|)|j\rangle$. In other words, I'm saying that studying $\int dU U^{\otimes t}\otimes (\bar U)^{\otimes t}$ amounts to studying "twirling channels" of the form $$\Phi_t(X) = \int dU\, U^{\otimes t}X (U^\dagger)^{\otimes t}.$$ In turn, the study of these channels generalises the study of objects like $\int d\psi \mathbb{P}_\psi^{\otimes t}$, because $$\Phi_t(\mathbb{P}_0^{\otimes t}) = \int dU\, \mathbb{P}(U|0\rangle)^{\otimes t} = \int d\psi \, \mathbb{P}_\psi^{\otimes t},$$ where I'm using the symmetry of the Haar-uniform integrals over states and unitaries to observe that picking $U|0\rangle$ uniformly at random over $U$ is the same as picking a general uniformly-random pure state $|\psi\rangle$. Also, the choice of $|0\rangle$ as reference state is here entirely arbitrary.
This tells you that studying the twirling the channels is more general than just studying the sum over projectors. To actually compute these quantities, you can either study the irreps of the appropriate tensor product representations of the unitary group, as discussed e.g. in Averaging over a single Haar-random unitary applied $t$ times, Random quantum states and Schur-Weyl duality, and links therein, or if you prefer a more "computational" approach, just decompose the expressions in components, and look up the known formulas for Weingarten functions.
Regarding the explicit case you mentioned, from the formulas in https://en.wikipedia.org/wiki/Weingarten_function you can derive the expression $$\Phi_2(X)\equiv \int dU\, U^{\otimes 2}X(U^\dagger)^{\otimes2} = \frac{1}{d^2-1}[\operatorname{tr}(X)I + \operatorname{tr}(WX)W] -\frac{1}{d(d^2-1)} [\operatorname{tr}(WX)I + \operatorname{tr}(X)W],$$ with $W\equiv \sum_{ij} |i,j\rangle\!\langle j,i|$ the swap operation. Therefore $$\Phi_2(\mathbb{P}_{00}) = \frac{I+W}{d(d+1)}.$$ Notice that $(I+W)/d(d+1)$ is the projection onto the symmetric subspace, compatibly with the statements above about group representations etc, and it also equals $\int d\psi\, \mathbb{P}_\psi^{\otimes2}$.
On the other hand, $\int dU \, U\otimes\bar U$ relates to the channel $$\Phi_1(X)=\int dU\, UXU^\dagger = \frac{\operatorname{tr}(X)I}{d}.$$ This can be written in more abstract form as $\Phi_1=|I/\sqrt d\rangle\!\rangle \langle\!\langle I/\sqrt d|$, meaning the projection onto the (normalised) operator $I/\sqrt d$. And a map $X\mapsto \langle A,X\rangle B$ has natural representation $|A\rangle\!\rangle \!\langle\!\langle B|$, from which we conclude that $$K(\Phi_1) = \int dU \, U\otimes\bar U = |m\rangle\!\langle m|,$$ with $|m\rangle\equiv\frac{1}{\sqrt d}\sum_i |i,i\rangle$ the maximally entangled state. Finally, you obtain the expression for $\int dU\, U\otimes U^\dagger$ by taking the partial transpose of this, and end up with the swap operator on the right-hand side.
