# Is the Haar measure invariant under conjugation?

Denote the Haar measure on the unitary group $$U(\mathcal X)$$ by $$\eta$$. Does this equation hold (assuming the integral exists):

$$\int d\eta(U) f(U) = \int d\eta(U) f(U^\dagger)$$?

Intuitively this makes sense because choosing a random $$U$$ seems to be the same as choosing a random $$U^\dagger$$, but I'm not sure how to prove this.

I will answer this question in a more general context. You might know that Haar's theorem tells you that on any locally compact group $$G$$, there is a unique left-invariant (Borel) measure $$\mu$$, up to a positive constant. Left-invariance means that $$\mu(g A)=\mu(A)$$ for any $$g\in G$$ and (measurable) set $$A\subset G$$. This is the (left-) Haar measure on $$G$$.
The same argument yields a unique right-invariant measure, up to a constant. Note that we can always turn a left-invariant Haar measure into a right-invariant Haar measure by taking the inverse: $$\tilde\mu(A) := \mu(A^{-1})$$. That is exactly what you are interested in.
For certain groups, the left- and right-invariant Haar measures coincide (unimodular groups) and the unitary group $$U(d)$$ is such a group. For these groups, your statement is correct. The proof is straightforward:
Unimodularity implies that the "inverse" of the normalised Haar measure $$\mu$$ on $$U(d)$$ is itself a left- and right-invariant Haar measure on $$U(d)$$. By uniqueness, $$\tilde\mu$$ can only differ by a positive constant from $$\mu$$. However, it is clear that both measures are normalised, thus $$\tilde\mu = \mu$$.