See the comments, What does the mut self_encoded means here?

pub trait DecodeLength {
    fn len(self_encoded: &[u8]) -> Result<usize, Error>;
}

// This can compile.
impl DecodeLength for i32 {
    // here
    fn len(mut self_encoded: &[u8]) -> Result<usize, Error> {
       usize::try_from(u32::from(Compact::<u32>::decode(&mut self_encoded)?))
            .map_err(|_| "Failed convert decoded size into usize.".into())
    }
}

//  This way not works since the signature of this len is not correctly.
// impl DecodeLength for i32 {
//     fn len(self_encoded: &mut [u8]) -> Result<usize, Error> {
//         Ok(2)
//     }
// }

>Solution :

you must remember when you change your fn like this

fn len(mut self_encoded: &[u8]) -> usize {
    2
}

you didn’t change actual input, your input still is &[u8], you just tell the compiler that value of input variable can be changed
like this

impl DecodeLength for i32 {
    // here
    fn len(mut self_encoded: &[u8]) -> usize {
        self_encoded = b"123";
        2
    }
}

but when you change the input type &[u8] to &mut [u8] you chose a new type for input

now compiler give you an error and says "expected &[u8] but found &mut [u8]

//  Error: expected fn pointer `fn(&[u8]) -> _` found fn pointer `fn(&mut [u8]) -> _
impl DecodeLength for i32 {
    fn len(self_encoded: &mut [u8]) -> usize {
        2
    }
}

remember &[u8] and &mut [u8] are different and have different use

[Edit]

you can’t use ? in function when your function doesn’t return Result

[Edit 2]

look at the following code

impl DecodeLength for i32 {
    // here
    fn len(mut self_encoded: &[u8]) -> usize {
      let mut_ref = &mut self_encoded; // this give you `&mut &[u8]`
      let mut_ref2 = self_encoded.as_mut(); // Error: `as_mut` is only valid for `&mut` references
      1
    }
}

you can’t change &[u8] type into &mut &[u8], this will give you an error

the simple way is change your DecodeLength like this

pub trait DecodeLength {
    fn len(self_encoded: &mut [u8]) -> Result<usize, Error>;
}