We know there is at least one gun in the ISS the caliber seems to a state secret.

According to Wikipedia and NASA large pdf there are over 100 shield configurations on the ISS. These would be optimized for external impact protection.

What is the best and worse case scenario for a gun fired inside the ISS?

Best case; Lowest energy gun fired at the most protected area of the hull

Worst case; Highest energy gun fired at the least protected area of the hull

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    $\begingroup$ Firing a gun inside a tin can is not usually recommended... $\endgroup$ – Deer Hunter May 27 '14 at 11:37
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    $\begingroup$ Firing a gun inside a $150 billion tin can hurtling through a total vacuum at 8 km/s is definitely not usually recommended. $\endgroup$ – Vedant Chandra May 27 '14 at 13:46
  • $\begingroup$ The bullet would be so much worse if it was fired from some sort of space elevator. The relative velocity of the collision would be around 7km/s. $\endgroup$ – ThePlanMan May 28 '14 at 16:25
  • $\begingroup$ @DeerHunter it's not a tin can, I believe it is mostly aluminium. $\endgroup$ – James Jenkins Nov 18 '15 at 15:22
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    $\begingroup$ @JamesJenkins most tin cans aren't tin either, typically steel or aluminium. $\endgroup$ – Baldrickk Nov 2 '17 at 14:07

There are lots of variables here. Gun type, wall thickness, type of shielding used.
According to this email exchange, handgun bullets can penetrate 1-3 cm of aluminium. It also states that:

(in) a M113A2 APC (armored personnel carrier) the aluminum (hull) is about 3/4 inches

The first data I've found for the wall thickness of a space station module is for ESA's proposed Columbus module: they used a Whipple shield with two walls of 2 mm and 3 mm thick. At first glance, not enough to stop a bullet. And the bullet will probably be going too slow to disintegrate on impact (as a micrometeoroid would).

As far as I know, in a Whipple shield, only the inner wall is airtight, so if you shoot from the inside, you're going to have a leak. At ~1 cm diameter, it's not big enough to cause immediate catastrophic decompression, but it'd be time for emergency procedures.

Worst case is that your bullet trajectory goes through a fuel (or oxidizer) tank. This will most likely result in contamination and exposure to highly toxic substances.

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    $\begingroup$ I think the point is: there is a misconception that firing any projectile weapon into the wall of any pressure vessel will cause the vessel to burst explosively. In other words, your little 9mm bullet will tear through the pressure wall, creating a rupture which almost instantaneously expands to metres in diameter and beyond, causing all of the vessel's contents to be ejected into space. The far less dramatic truth being that a 9mm bullet will puncture a hole just about 9mm in diameter, the worst consequence of which is that air will gradually leak out until it is plugged. $\endgroup$ – Anthony X May 28 '14 at 0:17
  • $\begingroup$ @AnthonyX would a cannon-ball sized hole be different? Explosive decompression at that size? What size would explosively decompress? $\endgroup$ – Magic Octopus Urn Nov 9 '18 at 17:54
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    $\begingroup$ @MagicOctopusUrn A key point in all of this relates to the way the skin material (an aluminum alloy) will react to an impact. Such materials have some degree of ductility, so they will undergo some amount of inelastic deformation before fracturing. That deformation absorbs energy locally and tends to keep the effect of something like a bullet impact from propagating. $\endgroup$ – Anthony X Nov 10 '18 at 0:30
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    $\begingroup$ @MagicOctopusUrn (I just ran across this while looking for ISS module wall thicknesses.) What Anthony X says is right—up to a point. There is a kind of material defect called a "Griffith crack" that will propagate catastrophically. When a material under tension—as a pressure vessel wall is—develops a crack, the edges of the crack pull apart, releasing some strain energy. The energy released is proportional to the square of the crack length. If the crack is large enough, the strain energy released is large enough to break more material, lengthening the crack, releasing more energy...BOOM!... $\endgroup$ – Tom Spilker Nov 13 '18 at 3:13
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    $\begingroup$ @MagicOctopusUrn A 9 mm bullet probably wouldn't create a Griffith crack. The edges of that bullet hole aren't crack-free. True, some of the material deforms plastically, but right at the edge the material is in failure, not plastic deformation, so it tends to be kind of ragged, and ragged-ness is what Griffith cracks feed off of. The distance from a crack on one side of the bullet hole to a similar crack on the other side is the dimension related to the Griffith criterion. A hole more like 30 cm (~1 US foot) might satisfy the Griffith criterion, and if so that crack would propagate. $\endgroup$ – Tom Spilker Nov 13 '18 at 3:21

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