Reducing the white spacing

With this simpler code, it can all fit on a single page. I loaded nccmath for its medium-sized fractions, which look better for coefficients, in my opinion:

\documentclass[11pt, a4paper]{report}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{bm}
\usepackage{nccmath}
\usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs}
\usepackage{color}
\usepackage{array}
\usepackage{setspace}% if you must (for double spacing thesis)
\usepackage{fancyhdr}
\usepackage{enumitem}
\usepackage{tikz}
\usepackage{parskip}
\usepackage{lipsum}
\usepackage{floatrow}

\begin{document}

\newcommand{\iu}{{i\mkern1mu}}
\[
\setlength\extrarowheight{3pt}
    \begin{array}{c | c c c c c }
    & 0 & 1 & 2 & 3 & 4\\
    \cline{1-6}
    \chi_0 & 1 & 1 & 1 & 1 & 1\\
    \chi_1 & 1 & a & a^2 & a^3 & a^4\\
    \chi_2 & 1 & a^2 & a^4 & a & a^3\\
    \chi_3 & 1 & a^3 & a & a^4 & a^2\\
    \chi_4 & 1 & a^4 & a^3 & a^2 & a\\
    \end{array}
\]

with $a = \exp\bigl\{\frac{2\pi \iu}{5}\bigr\}$, hence $a^5=1$ with $|G|=5$.

Applying the definition of Fourier transform from definition 3.1.2 we have:
    \begin{fleqn}
    \begin{align*}
        \hat{f}(\chi_0) & =f(0)+f(1)+f(2)+f(3)+f(4) \\
        \hat{f}(\chi_1) & =f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4) \\
        \hat{f}(\chi_2) & =f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4) \\
        \hat{f}(\chi_3) & =f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4) \\
        \hat{f}(\chi_4) & =f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)
    \end{align*}
    \end{fleqn}
        Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
\allowdisplaybreaks
        \begin{align*}
        {f}(0)
          &=\mfrac{1}{5}\bigl[ \hat{f}(\chi_0)+\hat{f}(\chi_1)+\hat{f}(\chi_2)+\hat{f}(\chi_3)+\hat{f}(\chi_4)\bigr]\\
          & = \begin{aligned}[t]
           &\mfrac{1}{5}\bigl[f(0)+f(1)+f(2)+f(3)+f(4)]\\
            & + \mfrac{1}{5}\bigl[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)\bigr]\\
            & + \mfrac{1}{5}\bigl[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)\bigr]\\
            & + \mfrac{1}{5}\bigl[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)\bigr]\\
            & + \mfrac{1}{5}\bigl[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)\bigr]
           \end{aligned}\\
      & =f(0) \begin{aligned}[t] 
            & + \mfrac{f(1)}{5}[1+a+a^2+a^3+a^4]\\
            & + \mfrac{f(2)}{5}[1+a+a^2+a^3+a^4]\\
            & + \mfrac{f(3)}{5}[1+a+a^2+a^3+a^4]\\
            & + \mfrac{f(4)}{5}[1+a+a^2+a^3+a^4]
            \end{aligned}\\
        & = f(0)
        \shortintertext{Similarly:}
        {f}(1)
          &= \mfrac{1}{5}\Bigl[\hat{f}(\chi_0)+\mfrac{1}{a}\hat{f}(\chi_1)+\mfrac{1}{a^2}\hat{f}(\chi_2)+\mfrac{1}{a^3}\hat{f}(\chi_3)+\mfrac{1}{a^4}\hat{f}(\chi_4)\Bigr]\\
             & = f(1) \\[1.5ex]
         f(2)
         &= \mfrac{1}{5}\bigl[\hat{f}(\chi_0)+a^2\hat{f}(\chi_1)+a^4\hat{f}(\chi_2)+a\hat{f}(\chi_3)+a^3\hat{f}(\chi_4)\bigr] \\
          & = f(2) \\[1.5ex]
        f(3)
         &= \mfrac{1}{5}\bigl[\hat{f}(\chi_0)+a^3\hat{f}(\chi_1)+a\hat{f}(\chi_2)+a^4\hat{f}(\chi_3)+a^2\hat{f}(\chi_4)\bigr] \\
         & = f(3) \\[1.5ex]
        f(4)
          &= \mfrac{1}{5}\bigl[\hat{f}(\chi_0)+a^4\hat{f}(\chi_1)+a^3\hat{f}(\chi_2)+a^2\hat{f}(\chi_3)+a\hat{f}(\chi_4)\bigr] \\
          & = f(4)
        \end{align*}

    \end{document} 

You should avoid \\ on the last line of alignments. Perhaps the following is closer to what you want:

\documentclass[11pt, a4paper]{report}

\usepackage{amsmath,array}

\begin{document}

\newcommand{\iu}{{i\mkern1mu}}
\begin{equation*}
  \setlength\extrarowheight{3pt}
  \begin{tabular}{c | c c c c c }
    & $0$ & $1$ & $2$ & $3$ & $4$\\
    \cline{1-6}
    $\chi_0$ & $1$ & $1$  &  $1$   & $1$   & $1$\\
    $\chi_1$ & $1$ & $a$  &  $a^2$ & $a^3$ & $a^4$\\
    $\chi_2$ & $1$ & $a^2$ & $a^4$ & $a$   & $a^3$\\
    $\chi_3$ & $1$ & $a^3$ & $a$   & $a^4$ & $a^2$\\
    $\chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\\
  \end{tabular}
\end{equation*}
with $a = \exp\{\frac{2\pi\iu}{5}\}$ hence $a^5=1$ with $|G|=5$.

Applying the definition of Fourier transform from Definition~3.1.2 we
have:
\begin{align*}
  \hat{f}(\chi_0) &=f(0)+f(1)+f(2)+f(3)+f(4),\\
  \hat{f}(\chi_1) &=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4),\\
  \hat{f}(\chi_2) &=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4),\\
  \hat{f}(\chi_3) &=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4),\\
  \hat{f}(\chi_4) &=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4).
\end{align*}
Using Definition~3.1.3 we can compute the inverse Fourier transform
$f(t)$:
\begin{align*}
  f(0)
  &=\frac{1}{5}[ \hat{f}(\chi_0) + \hat{f}(\chi_1) + \hat{f}(\chi_2) +
    \hat{f}(\chi_3) + \hat{f}(\chi_4)]\\
  &=\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]\\
  &\qquad + \frac{1}{5}[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\\
  &\qquad + \frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\\
  &\qquad + \frac{1}{5}[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\\
  &\qquad + \frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
  \\
  &= f(0)\\
  &\qquad + \frac{f(1)}{5}[1+a+a^2+a^3+a^4]\\
  &\qquad +\frac{f(2)}{5}[1+a+a^2+a^3+a^4]\\
  &\qquad +\frac{f(3)}{5}[1+a+a^2+a^3+a^4]\\
  &\qquad +\frac{f(4)}{5}[1+a+a^2+a^3+a^4]\\
  &=f(0).
\end{align*}
Similarly
\begin{align*}
  f(1)
  &= \frac{1}{5}\Bigl[\hat{f}(\chi_0) + \frac{1}{a}\hat{f}(\chi_1) +
    \frac{1}{a^2}\hat{f}(\chi_2) + \frac{1}{a^3}\hat{f}(\chi_3) +
    \frac{1}{a^4}\hat{f}(\chi_4)\Bigr]\\
  &=f(1),\\
  f(2)
  &= \frac{1}{5}[\hat{f}(\chi_0) + a^2\hat{f}(\chi_1) +
    a^4\hat{f}(\chi_2) + a\hat{f}(\chi_3) + a^3\hat{f}(\chi_4)]\\
  &=f(2), \\
  f(3)
  &= \frac{1}{5}[\hat{f}(\chi_0) + a^3\hat{f}(\chi_1) +
    a\hat{f}(\chi_2) + a^4\hat{f}(\chi_3) + a^2\hat{f}(\chi_4)]\\
  &=f(3),\\
  f(4)
  &= \frac{1}{5}[\hat{f}(\chi_0) + a^4\hat{f}(\chi_1) +
    a^3\hat{f}(\chi_2) + a^2\hat{f}(\chi_3) + a\hat{f}(\chi_4)]\\
  & =f(4).
\end{align*}
\end{document}