Let $R: U(1) \to \mathrm{GL}(n, \mathbb C)$ be a representation, let $e^{i\theta} \in U(1)$, and let $\langle \cdot , \cdot \rangle$ be the standard inner product on $\mathbb{C}^n$. Then, the Hermitian inner product on $\mathbb{C}^n$ $$\langle u,v \rangle_R := \frac{1}{2\pi}\int_0^{2\pi} \langle R(e^{i\theta})u,R(e^{i\theta})v \rangle \, \mathrm{d}\theta $$ is invariant under the action of $R$ on $\mathbb{C}^n$, meaning $\langle u,v \rangle_R = \langle R(e^{i\phi})u,R(e^{i\phi})v \rangle_R$.
Proof
It should be really easy to prove, but I am getting stuck on the last step. $$\langle R(e^{i\phi})u,R(e^{i\phi})v \rangle_R = \frac{1}{2\pi}\int_0^{2\pi} \langle R(e^{i\phi})R(e^{i\theta})u,R(e^{i\phi})R(e^{i\theta})v \rangle \, \mathrm{d}\theta = \frac{1}{2\pi}\int_0^{2\pi} \langle R(e^{i(\theta+\phi)})u,R(e^{i(\theta+\phi)})v \rangle \, \mathrm{d}\theta = \frac{1}{2\pi}\int_{\phi}^{\phi+2\pi} \langle R(e^{i\theta'})u,R(e^{i\theta'})v \rangle \, \mathrm{d}\theta'$$ Where I used the fact that $R$ is a representation and a change of variable $\theta' = \theta + \phi$. This is almost $\langle u, v \rangle_R$, except for the boundaries of integration shifted by $\phi$. Is this still equivalent to the original inner product? I should also note that I can't use the classification of the representations of $U(1)$ with $\mathrm{GL}(n, \mathbb C) \ni R(e^{i\theta}) = diag(e^{im_k\theta})_n$ with $k=1,...,n$ and $m_k \in \mathbb Z$ since this lemma is supposed to be used to prove it.
So you have the right idea, you just missed a crucial fact. $d\theta$ is a Haar measure, so it's almost by definition that
$$ \int_{[0,2\pi]} f(\theta+\phi) \,d\theta = \int_{[0,2\pi]} f(\theta)\,d\theta.$$
This trick works for any compact abelian group actually. Given a compact abelian group $G$ with Haar measure $dg$, Hilbert space $H$ with inner product $\langle\cdot,\cdot\rangle$, and representation $\rho$ of $G$ on $\mathcal{B}(H)$, the inner product
$$ \langle f_1,f_2\rangle_{\rho} = \int_G \langle \rho(g)f_1, \rho(g)f_2\rangle\,dg$$
is invariant under actions by $\rho$, i.e.
$$ \langle \rho(g')f_1,\rho(g')f_2\rangle_{\rho} = \langle f_1,f_2\rangle_{\rho}.$$
This lets you work with unitary representations. It's called a group averaging trick.