Fixed point conversions
As mentioned in the comments, your question is pretty confusing and confused, so I might not be able to answer all your smaller questions, but I'll try to shed some light on the topic. This should help you understand what's going on, and hopefully this understanding will help you answer all further questions by yourself.

In the following I assume two's complement representation. Note that if $M$ is the word length, and $n$ is the number of fractional bits, a number $x$ is represented as

$$x=K\cdot 2^{-n}\tag{1}$$

where $K$ is an integer is the range $[-2^{M-1},2^{M-1}-1]$. Now if you want to change the number of fractional bits to $m$, you have

$$x=L\cdot 2^{-m}\tag{2}$$

Combining (1) and (2) gives

$$L=K\cdot 2^{m-n}\tag{3}$$

If $m&gt;n$, $K$ needs to be left-shifted by $m-n$ bits to obtain $L$. This means that you lose range (i.e. the maximum number that can be represented becomes smaller), so there is a possibility of overflow. If $m&lt;n$, $L$ is obtained from $K$ by right-shifting $n-m$ bits, but for two's complement you need to replicate the sign bit, i.e. if the original sign bit was set, the new right-shifted number must also have the sign bit set. For $m&lt;n$ you can lose precision because the resolution is reduced.

An example makes things easier to see. Let $M=6$, i.e. the integer $K$ is in the range $[-32,31]$. Let the number of fractional bits be $n=3$. E.g.

$$x=\texttt{101.011}=(-2^{5}+2^{3}+2^{1}+2^{0})\cdot 2^{-3}=-21\cdot 2^{-3}=-2.625$$

If we want to change the number of fractional bits to $m=2$, you need to do an arithmetic right-shift, resulting in

$$y=\texttt{1101.01}=-11\cdot 2^{-2}=-2.75\neq x$$

i.e. precision is lost. If instead we want $m=4$ fractional bits we need to do a left-shift resulting in overflow:

$$y=\texttt{01.0110}=22\cdot 2^{-4}=1.375\neq x$$

This is clear because by increasing resolution we have decreased the range of possible numbers to $[-32,31]\cdot 2^{-4}=[-2,1.9375]$, which makes it impossible to represent the original value $x=-2.625$.

There are of course examples where you would neither lose precision nor would you get overflow. E.g. (again with $M=6$ and $n=3$)

$$x=\texttt{001.010}=10\cdot 2^{-3}=1.25$$

For $m=2$ we get

$$y=\texttt{0001.01}=5\cdot 2^{-2}=1.25 = x$$

i.e. no loss in precision, and for $m=4$ we get

$$y=\texttt{01.0100}=20\cdot 2^{-4}=1.25=x$$

i.e. no overflow.

I hope that this addressed at least some of your questions and that things have become a bit clearer. If you haven't done it yet, you should check out the basics here, here, and here.