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 homomorphism, 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 $m$.

The problem is that in the OP's situation $p_m$ is not surjective (unless $m=1$ or $n=1$) and the argument fails.

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 homomorphism¹, 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 $m$.

¹ 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...