Using gulp-imgmin is it bad to compress or minify an image more than one time?

I have set up the following gulpfile in order to test gulp-imagemin. I've included some additional imagemin plugins.

var gulp = require('gulp'),
imagemin = require('gulp-imagemin');

gulp.task('imagemin', function() {
    return gulp.src('src/lib/images/**.*')
           .pipe(imagemin({
                progressive: true
           }))
           .pipe(gulp.dest('src/lib/images/'))
});

gulp.task('default', ['imagemin']);

Here's my initial folder structure (only one image included)

src/ 
|-- lib/ 
|   |-- images/
|   |   |-- nasa.jpg

First run of gulp outputs the following

[22:31:32] gulp-imagemin: Minified 1 image (saved 489 B - 2.1%)

Rerunning gulp again outputs the following

[22:32:43] gulp-imagemin: Minified 1 image (saved 0 B - 0%)

---

To answer your questions

I am trying to decide if I should just compress/minify all my images and have them just overwrite each other, or if I should store the minified/compressed versions of each image in a separate folder?

As suggested by Charminbear, it's advised to split your workspace into for instance, 'src' (Source) and 'dist' (Distribution) folders. You can store your uncompressed images inside of src/lib/images and use gulp to serve your minified images to dist/lib/images.

src/ 
|-- lib/ 
|   |-- images/  
|   |   |-- nasa.jpg
|   |   |-- mars.jpg
dist/
|-- lib/
|   |-- images/
|   |   |-- nasa.jpg
|   |   |-- mars.jpg
gulpfile.js

What is the role of src and dist folders?

Or when gulp-imgmin goes to compress the image, does it see its already compressed and skip it?

gulp-imagemin will not try to minify already minified images.