Let us get things clear in our minds: 1. you are referring to a magnifying lens appearring inverted (NO, it is the IMAGE PRODUCED BY THE LENS that matters NOT the lens itself) 2. You mention a magnifying lens producing a diminished image - surely that is a self-contradiction.
Let me take through the optics of it all step by step.
Firstly, a magnifying lens is a convex or converging lens (its power is in plus dioptres [inversely related to its radius of curvature) i.e. it is thicker in the middle thin at the edges, therefore convex; it makes light rays converge i.e. bend towards its axis (= midline).
A diverging lens diverges rays = bends them (refracts) outwards away from the axis, and is thicker at the edges, thinner in the centre. (see pic below).
The image in diagram I (convex = magnifying lens), is inverted (A is on top in the object, and A' is below in the image) AND the image is on opposite side of the lens as the object, so can be seen where it is = real.
In diagram II, (concave = minifying lens), the image is on the same side of the lens as the object (= virtual image), is smaller (minifying) and is upright (A is above [object] AND A' is above [image]).
In the human eye, the crystalline lens (very convex [power = +60 dioptres! average magnifying lens in physics = 4-6 D]; almost like a marble!!) is of course a convex (magnifying) lens - the IMAGE ON THE RETINA is INVERTED yes, but we learn to re-invert it in neonatal life (when we are very young babies) [in fact, newborns see upside down in the first few hours/days].
In a camera the situation is very complex because this instrument [like many others - telescope, keratomter, slit lamp biomicroscope] uses a copound lens i.e. several lenses fused together inclusive of plus and minus ones. Since you seem to be a student with an inquiring mind, I will briefly mention to you the equation you need to work out the power of such lenses:
Attachment 639554639558Attachment 639554639558
FE = equivalent power; F1 =power of first lens; F2 = power ofsecond lens; d = distance between optical centres of the 2 lenses.
From FE and object details, you can work out size, position of image.
Very good Q btw!
M (Science tutor)