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}