Symmetric polynomials in variables $x_1, \dots, x_n$ are invariant under the action of $S_n$ which arbitrary permutes the variables.
Now, I'm reading that an invariant polynomial $f_{sym}$ can be obtained by any polynomial $f$ by symmetrization through group averaging.
$$f_{sym}(x_1, \dots, x_d) = \frac{1}{n!}\sum_{\rho \in S_n}f(x_{\rho(1)}, \dots, x_{\rho(n)})$$
where of course $n! = |S_n|$.
Now what I don't understand is why do we need the factor $\frac{1}{n!}$ as the summation should be enough to obtain such a polynomial.
EXAMPLE
$$f(x,y) = 4x^2+5y^4$$ $$f_{sym}(x,y) = 4y^4+5x^2+4x^2+5y^4$$
You need the factor only for one reason. If you take symmetric $f$ then should be $f_{sym} =f.$