Most generally, Schur's Lemma is used as a tool in representation theory. In this answer, I'll try to explain it without talking about said theory.
Preliminaries: Schur's Lemma
Let us consider a vector space $V$ on an algebraically closed field $\mathbb{K}$ (typically $\mathbb{C}$). We denote $\mathcal{L}(V)$ the space of endormophisms of $V$. Finally, let us consider a subspace $U$ of $\mathcal{L}(V)$.
$U$ is said to be irreducible if the only subspaces of $V$ that are left unchanged by every $u\in U$ are the nil subspace $\{0\}$ and $V$ itself. That is, we have:
$$\forall F\text{ subspace of }V, [\forall u\in U, \forall x\in F, u(x)\in F]\iff[F=\{0\}\lor F=E]$$
What Schur's Lemma tells you is that if an endomorphism $v$ commutes with all elements of an irreducible $U$, then $v$ must be an homothety, that is $v$ is proportional to the identity. That is:
$$[\forall u\in U, u\circ v=v\circ u]\implies[\exists\lambda\in\mathbb{K},v=\lambda\,\mathrm{id}]$$
Computing the integral
So now, let us consider your integral:
$$\tau=\int\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)$$
where $\mu$ is the Haar-measure on $H\otimes K$. Let us decompose the reasoning in two steps.
Showing that $\tau$ is an operator on the symmetric subspace
First of all, note that $\tau$ is an operator on the symmetric subspace $\mathrm{Sym}^n(H\otimes K)$. In order to show this, let us define $P_\pi$ as the permutation of the tensor factors according to the permutation $\pi\in\mathfrak{S}_n$. That is, if we have a state $\left|x_1,\cdots,x_n\right\rangle\in(H\otimes K)^{\otimes n}$, then:
$$P_\pi\left|x_1,\cdots,x_n\right\rangle=\left|x_{\pi^{-1}\left(1\right)},\cdots,x_{\pi^{-1}\left(n\right)}\right\rangle$$
Note that for any $\pi\in\mathfrak{S}_n$, we have:
$$P_\pi\tau=P_\pi\int\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)=\int P_\pi\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)=\int\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)=\tau$$
And similarly we have $\tau P_\pi=\tau$. The penultimate equation is more easily seen when writing $\sigma$ as $|\psi\rangle\!\langle\psi|$, since $\sigma^{\otimes n}=|\psi\rangle^{\otimes n}\!\langle\psi|^{\otimes n}$ and $P_\pi|\psi\rangle^{\otimes n}=|\psi\rangle^{\otimes n}$.
As a recall, the symmetric subspace is defined by the fact that $|\psi\rangle\in\mathrm{Sym}^n(H\otimes K)$ if and only if $P_\pi|\psi\rangle=|\psi\rangle$ for every $\pi\in\mathfrak{S}_n$.
Let us consider $|\psi\rangle\in\mathrm{Sym}^n(H\otimes K)$. We have for any $\pi\in\mathfrak{S}_n$:
$$P_\pi(\tau |\psi\rangle)=\left(P_\pi\tau\right)|\psi\rangle=\tau|\psi\rangle$$
Thus, $\mathrm{Sym}^n(H\otimes K)$ is stable by $\tau$. On the other hand, it is known that:
$$P=\frac{1}{n!}\sum_{\pi\in\mathfrak{S}_n}P_\pi$$
is an orthogonal projector on $\mathrm{Sym}^n(H\otimes K)$. This means that:
$$(H\otimes K)^{\otimes n}=\mathrm{Sym}^n(H\otimes K)\overset{\perp}{\oplus}\mathrm{Ker}(P)$$
So let us consider $|\psi\rangle\in\mathrm{Ker}(P)$. Note that $\tau=\tau P$. Thus:
$$\tau|\psi\rangle=\tau P|\psi\rangle=\tau0=0$$
Thus, $\tau$ is an operator on $\mathrm{Sym}^n(H\otimes K)$: we only have to determine its behavior on this space to know it completly.
Using Schur's Lemma
We know want to find an irreducible subspace of operators on $\mathrm{Sym}^n(H\otimes K)$ such that $\tau$ commutes with every operator in this subspace. Note that, denoting $\mathcal{U}$ the space of unitaries on $H\otimes K$:
$$\begin{align}
\forall U\in\mathcal{U},U^{\otimes n}\tau&=\int(U|\psi\rangle)^{\otimes n}\langle\psi|^{\otimes n}\,\mathrm{d}\mu(|\psi\rangle)\\
&=\int|\psi\rangle^{\otimes n}(\langle\psi|U)^{\otimes n}\,\mathrm{d}\mu\left(U^\dagger|\psi\rangle\right)\\
&=\int|\psi\rangle^{\otimes n}(\langle\psi|U)^{\otimes n}\,\mathrm{d}\mu(|\psi\rangle)\\
&=\int|\psi\rangle^{\otimes n}\!\langle\psi|^{\otimes n}\,\mathrm{d}\mu(|\psi\rangle)U^{\otimes n}\\
&=\tau U^{\otimes n}\end{align}$$
Thus, $\tau$ commutes with every element of $\left\{U^{\otimes n}\middle|U\in\mathcal{U}\right\}$. It has been proven that this space is irreducible with respect to $\mathrm{Sym}^n(H\otimes K)$. We can thus apply Schur's Lemma, which gives us that $\tau$ acts as the identity (up to a multiplicative factor) on $\mathrm{Sym}^n(H\otimes K)$. Since it is nil outside of this space, we actually have:
$$\exists\lambda\in\mathbb{C},\tau=\lambda P$$
We now only have to determine $\lambda$. But we know that $\mathrm{tr}(\tau)=1$ since it's a density matrix. Thus:
$$1=\lambda\mathrm{tr}(P)$$
Since $P$ is a projector on $\mathrm{Sym}^n(H\otimes K)$, its trace is equal to the dimension of $\mathrm{Sym}^n(H\otimes K)$, which is known to be $\binom{\mathrm{dim}(H\otimes K) + n - 1}{n-1}$. This finally gives us:
$$\tau=\frac{1}{\binom{\mathrm{dim}(H\otimes K) + n - 1}{n-1}}P$$
