I want to find the answer of 
$$\int dU \ U^m X \ U^{\dagger m}$$


Where $m\in\mathbb{N}$ and $U$'s are unitary matrices in $U(n)$ and $dU$ is a normalized Haar measure. $X$ is a given self-adjoint matrix. 


I used Schur lemma and found that the answer is of the form


*(Edit: the correct form is )*
$$pX+(1-p)tr(X)\frac{I}{n}$$


with $p\in[0,1]$, but if this is true then by comparing above expressions and using the results obtained in [my last question](https://mathoverflow.net/questions/195186/expectation-of-trace-of-nth-power-of-unitary-matrices), $p$ can be found as
$$p=\frac{m-1}{n^2-1}$$
which is greater than 1 for $m>n^2$.


I don't know what is going wrong here. Does anyone have an idea about my solution or another way for solving this integral?



An expectation of the product of random unitaries
Let me try to work this out, along the lines of a [similar calculation](http://mathoverflow.net/questions/194840/some-calculus-in-the-orthogonal-group-on/194900#194900) in the orthogonal (rather than unitary) group.


We need the fourth-order tensor
$$\int\_{{\rm U}(n)}(U^m)\_{ij}(\bar{U}^m)\_{kl}\,dU=a\_{m}(n)\delta\_{ij}\delta\_{kl}+b\_{m}(n)\delta\_{ik}\delta\_{jl}+c\_{m}(n)\delta\_{il}\delta\_{jk}$$
so that the required integral takes the form
$$\int\_{{\rm U}(n)}U^m X(U^\dagger)^m\,dU=a\_{m}(n)X+b\_{m}(n)\mathbb{1}\,{\rm tr}\,X+c\_{m}(n)X^{\rm t}
$$
*[note: The OP does not have the transpose $X^{\rm t}$, but I don't see a priori why this term will not appear.]*


Substitution $X=\mathbb{1}$ gives a first relation
$$a\_m(n)+nb\_m(n)+c\_m(n)=1$$
one more relation follows from application of theorem 2.1.b of [Diaconis and Evans](http://statweb.stanford.edu/~cgates/PERSI/papers/functionals.pdf):
$$n^2 a\_m(n)+nb\_m(n)+nc\_m(n)=\int\_{{\rm U}(n)}\,({\rm tr}\,U^m)({\rm tr}\,\bar{U}^m)\,dU={\rm min}\,(n,m)$$
*[note: in a related [MO posting](http://mathoverflow.net/questions/195186/expectation-of-trace-of-nth-power-of-unitary-matrices) I had $m$ instead of ${\rm min}\,(n,m)$, I have now corrected this oversight and apologize for the confusion it may have caused]*


I need a third relation
$$na\_m(n)+nb\_m(n)+n^2 c\_m(n)=\int\_{{\rm U}(n)}\,{\rm tr}\,(U^m\bar{U}^{m})\,dU$$


To evaluate this integral I write $U^m=V\Lambda^m V^\dagger$, with $\Lambda$ the diagonal matrix of eigenvalues of $U$ and $V\in{\rm U}(n)$ independent of $\Lambda$. I then first average over the $V$ matrices, which is easy because there are just four of them:
$$\int\_{{\rm U}(n)}\,{\rm tr}\,(V\Lambda^m V^\dagger\overline{V\Lambda^m V^\dagger})\,dV=\frac{1}{n+1}\left[({\rm tr}\,\Lambda^m)({\rm tr}\,\bar{\Lambda}^m)+{\rm tr}\,(\Lambda^m\bar{\Lambda}^m)\right]=\frac{1}{n+1}\left[({\rm tr}\,U^m)({\rm tr}\,\bar{U}^m)+n\right].$$
 and then the remaining average can be evaluated using Diaconis and Evans:


$$\int\_{{\rm U}(n)}\,{\rm tr}\,(U^m\bar{U}^{m})\,dU =\frac{n+{\rm min}\,(n,m)}{n+1}$$


so now we have three equations with three unknowns and we're done:


$$a\_{m}(n)= \frac{\min(n,m) -1}{n^2-1}=1-nb\_{m}(n),\;\;c\_m(n)= 0$$


and $c\_m(n)$ does in fact turn out to be equal to zero.


