For a funcion $\phi\in L_2$ let $\widehat{\phi}$ denote its Fourier transform. Denote the $n$-th Hermite function by $\phi_n$. We know three things:
1) By Fourier transform of the n-th derivative, $$\widehat{\frac{d^n\phi}{dx^n}}(w) = (iw)^n\widehat{f}(w)$$
2) By definition, $$\phi_n = \frac{d^n}{dx^n}e^{x^2}$$
3) Hermite functions are eigenvectors of the Fourier transform, specifically $$\widehat{\phi_n} = (-i)^n \phi_n.$$
But therefore [using 3) 2) 1) and 2) in order] $$(-i)^n \phi_n(x) = \widehat{\phi_n}(x) = \widehat{\frac{d^n}{dx^n}e^{x^2}} = (ix)^ne^{x^2} = (ix)^n\phi_0$$ [with a normalization constant ommited in the fourth term] and thus $$(-1)^n\phi_n(x) = x^n\phi_0(x)$$ which is clearly a contradiction, because of 2) for $n\geq 1.$
Do you have any idea of where is the error in my reasoning? is this an inconsistency in mathematics?
You mean $$\phi_n(x) = e^{x^2/2}\frac{d^n}{dx^n}e^{-x^2}$$ they are eigenfunctions of the Fourier transform $f\to \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i\omega x}dx$ (the constant factor is to make it unitary). The eigenvalue is automatically $\pm 1,\pm i$ because $\mathcal{F}^4 f=f$.