Is there a fixed-point free involution of the Möbius strip? Is there a fixed-point free continuous involution of the Möbius strip? (Meaning a function $f:M\to M$ such that $f\circ f={\rm id}$ and $f(x)\ne x$ for all $x$.)
The Lefschetz fixed-point theorem is useless here because the Möbius band is homotopic to a circle, which does have fixed-point free involutions. Beyond that, I don't know what tools to use.
(You can't just rotate it 180 degrees because doing that twice flips it along the other axis)
 A: This is essentially the argument Johannes Pauling mentioned in the comments, with a lot more detail.
First, since $f$ is a fixed-point-free involution, we really have a free action of $\mathbb{Z}/2\mathbb{Z}$ on $M$.  Thus, the quotient space $N:= M/(\mathbb{Z}/2\mathbb{Z})$ is naturally a manifold.
Proposition 1:  $N$ is non-orientable.
Sketch of Proof:   We show that more generally, given any covering map $X\rightarrow Y$ of manifolds, if $Y$ is orientable then so is $X$.  The idea is to pick a collection of charts witnessing the orientability of $Y$.  By restricting these charts to small enough open subsets (more specifically, by intersecting them with open sets in an evenly covering neighborhood of $Y$), we obtain a new cover of $Y$ by charts which a) still witness the orientability of $Y$ and b) all lie in an evenly covered neighborhood of the covering $X\rightarrow Y$.
Then pulling these charts back along the covering projection, we obtain disjoint copies of the charts on $X$.  It's easy to convince yourself that these charts on $X$ witness the orientability of $X$., because any overlap is a homeomorphic image of an overlap of charts on $Y$.  $\square$
Before we move on, let me note that it is not clear to me whether $M$ denotes the open Mobius band or the compact Mobius band with boundary, so I'll supply a proof for both cases.
Proposition 2A:  If $M$ is the Mobius band with boundary, then $M$ is homeomorphic to $N$.
Sketch:  Since $M$ has a boundary, then so does $N$.  In fact, since the boundary of $M$ is connected, the same must be true of $N$.  Also, Euler characteristic is multiplicative in coverings, so $\chi(N) = 0$.  Lastly, Proposition 1 tells us that $N$ is non-orientable.  According to wikipedia, this is enough to assert that $N$ is homeomorphic to $M$.  $\square$
Proposition 2B:  If $M$ is the open Mobius band, then $M$ is homeomorphic to $N$.
Sketch:  Using the double cover $M\rightarrow N$, the transfer homomorphism implies $H_1(N;\mathbb{Q})$ is at most one dimensional, since the same is true of $M$.  If $M$ is open, then so in $N$, and so $\pi_1(N)$ is free on some possibly infinite number of generators (see this MO question).  The Hurewicz theorem now gives that $H_1(N;\mathbb{Z})$ is a sum of $\mathbb{Z}$s, one for each generator, and the universal coefficients says the same thing for rational homology.  In particular, the number of generators is at most $1$, so we must have $\pi_1(N) \cong \mathbb{Z}$.  Thus, $H_1(N)\cong \mathbb{Z}$ as well.
According to this paper, we may obtain $N$ by starting with $S^2$, deleting points and disks, and then gluing the boundaries of the disks together.  Notice that deleting two disks and gluing the boundaries together in an orientation preserving way corresponds to connect summing with a torus.  In particular, this operation contributes $\mathbb{Z}^2$ to $H_1(M)$.  Since $H_1(N)\cong \mathbb{Z}$, we cannot use this operation.
Deleting two disks and gluing the boundary together by an orientation reversing map corresponds to connect summing with a Klein bottle.  Again, this creates too much homology.
If there is are any deleted disks, the boundary of each must be identified with itself.  The only way to do this can get a manifold is using a double covering of the boundary, so this process is essentially adding a cross cap (that is, connect summing with $\mathbb{R}P^2$.  If we do this two or more times, we have created too much homology (since $\mathbb{R}P^2 \sharp \mathbb{R}P^2$ is a Klein bottle), so we can only do this at most once.  On the other hand, if we don't do this, the only operation we have left is deleting points, which will give an orientable manifold.
Thus, $N$ is obtained by connecting summing $S^2$ with precisely one $\mathbb{R}P^2$ and then deleting points.  Lastly, if we delete more than one point, then there is once again too much homology, so $N$ is obtained by deleting a single point from $\mathbb{R}P^2$.  In other words, $N$ is an open Mobius band.  $\square$
Using Proposition $2$ to identify $N$ and $M$, we have found a double covering $M\rightarrow M$.  The next proposition says that the only double coverings of $M$ are orientable, giving our final contradiction.
Proposition 3:  There is, up to homeomorphism, a unique connected double covering of $M$.   This double covering is orientable.
Sketch of Proof:  If $M$ is open, then clearly $\mathbb{R}\times S^1$ double covers $M$.  If $M$ has boundary, then clearly $[0,1]\times S^1$ double covers $M$.  Both of these covers are orientable.  So, it's enough to show they are unique up to homeomorphism.  Covering space theory tells us that connected double covers of $M$ corresponds with an index $2$ subgroup of $\pi_1(M)\cong \mathbb{Z}$.  This group has a unique index $2$ subgroup ($2\mathbb{Z}$), so uniqueness follows. $\square$
