This is an elaboration on my comment.

Let's start in a more general setting since the comment was not about the groups $U(n)$:
Let $G$ be a compact group and $H$ a normalized Haar measure on it.
Let $p_m:G\to G$ be defined by $p_m(x)=x^m$, $m\in\mathbb Z$.
Now for any continuous function $f:G\to\mathbb R$ we have
$$
\int_Gf(p_m(x))dH(x)
=
\int_Gf(x)d(p_m)_*H(x).
$$
Since $p_m$ is a <del>homomorphism</del><sup>1</sup>, the pushforward $(p_m)_*H$ is a normalized Haar measure on $p_m(G)$.
If $p_m$ is surjective, it follows from the uniqueness of Haar measures that $p_{m*}H=H$ and thus
$$
\int_Gf(p_m(x))dH(x)
=
\int_Gf(x)dH(x).
$$
This idea made me think that the integral would be independent of $m$.

<hr>

<sup>1</sup>
The map is not a homomorphism.
The argument works (I think) for homomorphisms, but typically $p_m$ is not a homomorphism in a nonabelian group.
If $p_m$ is not a homomorphism, the pushforward doesn't have to be a Haar measure and the proof falls apart.

Maybe I'll let this answer stay here as a warning example...