Does bit-shift depend on endianness?

Suppose I have the number 'numb'=1025 [00000000 00000000 00000100 00000001] represented:

On Little-Endian Machine:

00000001 00000100 00000000 00000000

On Big-Endian Machine:

00000000 00000000 00000100 00000001

Now, if I apply Left Shift on 10 bits (i.e.: numb <<= 10), I should have:

[A] On Little-Endian Machine:

As I noticed in GDB, Little Endian does the Left Shift in 3 steps: [I have shown '3' Steps to better understand the processing only]

1. Treat the no. in Big-Endian Convention:

   00000000        00000000        00000100    00000001
   

2. Apply Left-Shift:

   00000000        00010000        00000100        00000000
   

3. Represent the Result again in Little-Endian:

   00000000        00000100        00010000        00000000 
   

[B]. On Big-Endian Machine:

00000000        00010000        00000100        00000000

My Question is:

If I directly apply a Left Shift on the Little Endian Convention, it should give:

numb:

00000001 00000100 00000000 00000000

numb << 10:

00010000 00000000 00000000 00000000

But actually, it gives:

00000000        00000100        00010000        00000000 

To achieve the second result only, I have shown three hypothetical steps above.

Please explain me why the above two results are different: The actual outcome of numb << 10 is different than the expected outcome.