Let $\mathcal{H}_N$ be the $N$ particle Hilbert space. So a quantum state $\left| \Psi \right>$ may be representated by
$$\left| \Psi \right> = \left| k_1 \right>^{(1)}\left| k_2 \right>^{(2)}…\left| k_N \right>^{(N)}$$
, where the $\left| k_i \right>^{(n)}$ each form a basis of the single particle Hilbert space. In order to obtain the fully symmetric subspace $\mathcal{H}_N^S$ the permutation Opertaror
$$\hat P \left[ \left| k_1 \right>^{(1)}\left| k_2 \right>^{(2)}…\left| k_N \right>^{(N)} \right] = \left| k_{P_1} \right>^{(1)}\left| k_{P_2} \right>^{(2)}…\left| k_{P_N} \right>^{(N)}$$
and the symmetrisation Operator
$$ \hat S = \frac{1}{N!} \sum_P \hat P$$
are introduced.
I am asked to prove that the symmetrisation Operator is hermitian. My first idea was that the permutation operator is hermitian and therefore the symmetrisation Operator, as a sum of hermitian Operators, is too. However I strugle to prove it and by now am unsure if $\hat P$ is hermitian after all. Any help is most welcome.
Edit
With some inspiration from the answers I think I got a prove. I think it is different from what the answers are aiming at, but I don't see how to do it in another way.
So here my shot: Every permutation Operator can be decomposed into a number of Exchange Operators $\hat E_{ij}$ that exchanges two particles:
$$\hat P = \prod_k \hat E_{i_k j_k}$$, with
$$\hat E_{ij} \left| k_1 \right>^{(1)}…\left| k_i \right>^{(i)}…\left| k_j \right>^{(j)}…\left| k_N \right>^{(N)} = \left| k_1 \right>^{(1)}…\left| k_{j} \right>^{(i)}…\left| k_{i} \right>^{(j)}…\left| k_N \right>^{(N)}$$
To prove that $\hat P$ (and therefore $\hat S$) is hermitian it is sufficient to prove that $\hat E_{ij}$ is. This finally can be done explicetly via
$$\left< k_1' \right|^{(1)}…\left< k_i' \right|^{(i)}…\left< k_j' \right|^{(j)}…\left< k_N' \right|^{(N)} \hat E_{ij} \left| k_1 \right>^{(1)}…\left| k_i \right>^{(i)}…\left| k_j \right>^{(j)}…\left| k_N \right>^{(N)}\\
=\delta(k_1',k_1)… \delta(k_i', k_j)… \delta(k_j', k_i)…\delta(k_N', k_N) $$
,which is the same as
$$\left< k_1 \right|^{(1)}…\left< k_i \right|^{(i)}…\left< k_j \right|^{(j)}…\left< k_N \right|^{(N)} \hat E_{ij} \left| k_1' \right>^{(1)}…\left| k_i' \right>^{(i)}…\left| k_j' \right>^{(j)}…\left| k_N' \right>^{(N)}\\
=\delta(k_1,k_1')… \delta(k_i, k_j')… \delta(k_j, k_i')…\delta(k_N, k_N') $$
Since both terms are reals, we may freely take the complex conjugate and arrive at the hermitian condition for the matrix elemets. I think this should prove it for $\hat E_{ij} \Rightarrow \hat P \Rightarrow \hat S$.
Edit 2
Darn, the prove doesn't work since the $\hat E_{ij}$ don't commute.
Best Answer
You may be getting dragged down by your notation. Why not consider the simplest possible case, with only two particles? Essentially, then, you have the two-particle basis $\newcommand{\ket}[1]{|#1\rangle}\{\ket{k_1}\ket{k_2}\}$, and a single permutation operator that acts as $$ \hat{P}\ket{k_1}\ket{k_2}=\ket{k_2}\ket{k_1}. $$ Can you use this to compute the matrix elements $\langle k_1'k_2'|\hat{P}|k_1k_2\rangle$ and $\langle k_1k_2|\hat{P}|k_1'k_2'\rangle$? How should they be related for $\hat{P}$ to be hermitian?
The general case is relatively similar. The permutation operator $\hat{P}$ corresponding to some permutation $P$ acts on basis states as $$ \hat{P}\ket{k_1\cdots k_N}=\ket{k_{P(1)}\cdots k_{P(N)}}. $$ This does not reduce to the two-electron case but an exactly analogous argument holds.