Why the singular values have to appear in descending order across the diagonal matrix? This question is about Singular Value Decomposition.
Given an arbitrary matrix $A$
$$A \in \mathbb{R}^{m\times n} $$
Its reduced SVD form is:-
$$A = U\Sigma V^T$$
Where as $\Sigma$ is the diagonal matrix containing scaling factors across its diagonal:-
$$ \Sigma =  \left( \begin{array}{ccccc}
\sigma_1 & \hfill & \hfill & \hfill & \hfill \\
\hfill & \sigma_2 & \hfill & \hfill & \hfill \\
\hfill & \hfill & \ddots &\hfill & \hfill \\
\hfill & \hfill & \hfill & \hfill & \sigma_n \\
\end{array} \right)$$
Now, it is said that the singular values across the diagonal are such that:-
$$\sigma_1 \geq \sigma_2 \geq \sigma_3 \geq \cdots \geq \sigma_n \geq 0 $$
Why do these values have to be necessarily in descending order across the diagonal? I read that it's a convention to write so, but there is more to the order than just convention. If this order is changed, we will get a totally different matrix instead of $A$. So the order is important. 
Or is it so that there is always guaranteed to be one unique solution for SVD in which these singular values across $\Sigma$ are guaranteed to be in descending order?

It is just convention.  If you change the order, and permute the columns of $U$ and $V$ correspondingly, you do get the same matrix $A$.  
