What can the fixed point spaces of an involution on the Klein bottle be? Suppose that I have a smooth involution acting on a Klein bottle. Then, what are the constraints on its set $F$ of fixed points? 
The first guess is to use Lefschetz formula for Euler's characteristic of $F$. However, this formula doesn't detect submanifolds in $F$ of Euler characteristic 0. So, for example, I can ''add'' to $F$ any bunch of disjoint circles, which seems quite counterintuitive.
 A: $F$ can be either one circle, one circle and two points, two circles, no points, or two points (or obviously the whole Klein bottle). Each of these corresponds to a single isomorphism class of involution. I construct and classify them below.
If you're only interested in obtaining what the possible fixed point spaces are, we can do that with equivariant cohomology. Smith theory says that for any smooth involution on a compact smooth manifold, $\dim H^*(M;\Bbb Z/2) \geq \dim H^*(F;\Bbb Z/2)$, where here I'm taking the total dimension (sum of Betti numbers). So $F$ is a union of circles and points, restricting us to either two circles, one circle and at most 2 points, or at most 4 fixed points. Of course, this leaves out the cases of 1, 3, and 4 fixed points (which can be dealt with by Lefschetz's fixed point theorem: because $\iota$ is a homeomorphism and the index of each fixed point is 1, there has to be either $0$ or $2$ isolated fixed points) or 1 circle and 1 point, in which case delete the circle; if the complement of a small open neighborhood is connected, it has rational Betti numbers $b_0 = b_1 = 1$ and all others zero, so must have zero or two fixed points by the same logic; if it's disconnected, the involution swaps the components and has no more fixed points.
Below is a "by hands" classification of the involutions of the Klein bottle which never invokes Smith theory (which involved spectral sequences and equivariant cohomology so may be more input data than you'd like.) 
I will occasionally in here say that there is only one or two types of involutions on a certain manifold; if left unjustified, these are not difficult.
Suppose $F$ contains a circle. The involution must act on the normal bundle of the circle by negation (or else the whole manifold would be fixed). If the circle's complement was disconnected, then the involution would identify the two components; so the involution must be isomorphic to the involution swapping two components of $\Bbb{RP}^2 \# \Bbb{RP}^2$ (since this is the only connected sum decomposition of the Klein bottle with both components homeomorphic). This is equivalent to saying that, given the presentation of the Klein bottle as the double of the Mobius band (take two copies and glue them together by the identity along the boundary), your involution must just swap the two bands.
Suppose the circle's complement is connected, and that the normal bundle is trivial. Then the involution restricts to an involution of the complement (which we may compactify by adding two different circles along the two different ends); Euler characteristic considerations show that the resulting object is the cylinder $S^1 \times [-1,1]$, and your involution swaps the two boundary circles, with $\iota(z, -1) = (\bar{z},1)$. (It must do this to descend to the Klein bottle.) To extend this to the whole cylinder, set $\iota(z,t) = (\bar{z},-t)$; this has two further fixed points. Indeed, this is up to isomorphism the only extension, as you can see by some fiddling with Lefschetz and equivariant CW decompositions (it cannot fix a circle because if so you could prove that $z$ and $\bar z$ are homotopic, and then it must have two fixed points, etc). (This is what you get when you let the negation involution of $\Bbb R^2/\Bbb Z^2$ descend to the Klein bottle.)
Now assume the normal bundle is nontrivial. Deleting a small invariant neighborhood of the circle leaves a Mobius band; so this involution is what we get when we glue two involutions of a Mobius band together. But an involution of the Mobius band (thought of as a $[-1,1]$ bundle over the circle) is necessarily either the identity or negation on the fiber or base (so there are four total). So your example is what you get when you glue two of the latter kind together.
So we see that if $F$ has a circle, the involution is either what you get when you swap the components of a connected sum decomposition $\Bbb{RP}^2 \# \Bbb{RP}^2$, in which case the fixed point set is a circle, or acting by a nontrivial involution in each factor of that connected sum (the connected sum circle invariant but not fixed under the involution). In this case $F$ is two circles. In all cases, if $F$ contains a circle, then it has no isolated fixed points.
Now suppose the fixed point set is 0-dimensional. There is a fixed point free involution of the Klein bottle (with quotient the Klein bottle). Presenting the Klein bottle as $S^1 \times \Bbb R/(\theta, t) \sim (-\theta, t+1)$, just send $(\theta, t) \to (i\theta, t+1/2)$.
Suppose the fixed point set is nonempty. Then the involution acts by negation on the tangent space of each fixed point, and the fixed point index of each is $1$. So we see that the Lefschetz number is the number of fixed points, and hence an involution with isolated fixed points must act by $-1$ on $H_1(K;\Bbb Q)$, and have precisely two fixed points. Deleting small invariant neighborhoods of each point, we get a fixed point free involution on $N_{2,2}$, the genus 2 nonoriented surface with two boundary components, that preserves (aka does not swap) the boundary components. $\chi(N_{2,2}) = -2$, so the quotient must be $N_{1,2}$ (since it will have 2 boundary components). You can classify by hand the connected double covers of $N_{1,2}$: they are precisely $\Sigma_{0,4}$ (the oriented double cover), $N_{1,3}$, and $N_{2,2}$. So there is actually a fixed point free involution upstairs extending the one we chose on the boundary (unique up to isomorphism) and thus there is a two-fixed-point involution of the Klein bottle.
