Let the needle's endpoints initially lie at two adjacent lattice points in the cartesian coordinate plane. It is clear that after some rotations, an endpoint's coordinates will be of the form: $(a+b\cdot 2^{-\frac{1}{2}}, c+d\cdot 2^{-\frac{1}{2}})$, where $a,b,c,d\in \mathbb Z$ Since, $1, 2^{-\frac{1}{2}}$ are linearly independent over $\mathbb Z$, this is a unique representation. the parity of $a+b$ is invariant for both endpoints. They start at different parity. So they can't interchange positions.

Solution from Twitch Solves ISL: The answer is no. Work in ${\mathbb Z}[\omega]$ where $\omega = \cos(45^{\circ})+i\sin(45^{\circ})$. Draw the needle as a directed segment from $0$ to $1$ in the plane. We will only keep track of the left end point: if the endpoint is located at $z$. Rotations around the other endpoint correspond to \[ z \mapsto z + \omega^k - \omega^{k-1} \]for some choice of $\omega$. The claim is that we never can reach $1$. To prove this we only need show the following claim, which proves the relevant invariant. Claim: $0 \not\equiv 1 \pmod{\omega-1}$ in ${\mathbb Z}[\omega]$. Proof. It suffices to show ${\mathbb Z}[\omega]/(\omega-1)$ is not trivial. Write ${\mathbb Z}[\omega] = {\mathbb Z}[T] / (T^4+1)$, then \[ {\mathbb Z}[\omega]/(\omega-1) \cong {\mathbb Z}[T] / \left( T^4+1, T-1 \right) \cong {\mathbb Z} / 2 = {\mathbb F}_2 \]as desired. $\blacksquare$

Essentially the same as #2, but this is how I thought about it Position the needle such that one endpoint is at $(0,0)$ and the other is at $(1,0)$, and WLOG rotate about the left endpoint first. Suppose a sequence of rotations works. Right before we switch endpoints to rotate around, draw the needle's current position on the plane. Also draw the line joining $(0,0)$ and $(1,0)$. Then the drawn segments clearly form a cycle/polygon. Furthermore, because the endpoints must be interchanged, there must be an odd number of vertices on this graph. On the other hand, every drawn segment has length $1$ and is either horizontal, vertical or has slope $\pm 1$. Because $1$ and $\sqrt{2}$ are linearly independent over $\mathbb{Z}$, it then follows that to end at the same position we started, we need an even number of horizontal and an even number of vertical edges. By rotating the argument $45^\circ$ the same is true for edges with slope $1$ and edges with slope $-1$, so there are an even number of edges and thus vertices: contradiction. $\blacksquare$

The answer should be no. Let $\omega = e^{\frac{\pi i}{4}}$. We start at $0$. A move consists of taking a complex number $z$ and adding $\omega^k$ where $k$ is some integer from $0$ to $7$. (Effectively what this does is that we rotate the needle by some amount, and then jump to the other endpoint.) We now show that the number of moves must be even, if we return to $0$. More formally, if \[ \sum_{k = 0}^{7} a_k\omega^k = 0 \]for some integers $a_0$ through $a_7$, we must have that \[ \sum_{k = 0}^{7} a_k \equiv 0 \pmod{2}.\] In fact we can assume that $a_4$ through $a_7$ are zero, by subtracting pairs of $0 = \omega^k + \omega^{k + 4}$ while preserving the parity. Now consider the real part of this thing. We have $a_0 + \frac{\sqrt{2}}{2}a_1 - \frac{\sqrt{2}}{2}a_3 = 0$. As such $a_0 = 0$. Likewise, $a_2 = 0$. This then forces $a_1 = a_3 = 0$, so the statement is true.

Does this work? The answer is no. Let $\omega$ be a primitive eighth root of unity. Suppose otherwise, and consider the locus of all positions of the segment. Modulo the first and last positions, all such segments $\ell$ form an equilateral polygon with an odd number of sides. As a result, there must exist some odd number of $\omega^k$'s that sum to $0$; equivalently, there exists a polynomial $f \in \mathbb Z[X]$ such that $f(1)$ is odd and $f(\omega) = 0$. But then $X^4+1 \mid f$, and hence $2 \mid f(1)$, contradiction!

I have a very silly solution, which by the looks of things looks kinda different from the other solutions. We place the needle on the complex plane with one end at $0$ and the other at $1$. We will let $z_0=0$ and $z_1=1$ and after $n$ moves, we let $z_n$ denote the new endpoint of the needle after a move. Note that if $\omega = e^{\frac{\pi i}{4}}$ then we can set up the following recurrence relation for $z_1,z_2, \ldots$ \[ z_{n} = \omega^{e_{n}}(z_{n-2}-z_{n-1})-z_{n-2} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2}\]for an arbitrary sequence of integers $e_1,e_2, \ldots$. This recurrence works because we translate $z_n$ to a new point $w$ on the unit circle, rotate it by $e_{n} \cdot \frac{\pi}{4}$ and undo the translation, which is exactly the operation we desire. Claim: If $k$ is odd and $a_1,a_2, \ldots, a_k$ are integers, then \[ \omega^{a_1}+\omega^{a_2}+ \cdots + \omega^{a_k} \not = 0 \] This is equivalent to showing that if $t_1\omega^1+t_2\omega^2+ \cdots +t_8\omega^8 = 0$ then $t_1+t_2+ \cdots +t_8$ is even. Indeed, note that \[ \Re(t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = t_1\cdot \Re(\omega^1)+t_2 \cdot \Re(\omega^2) \cdots +t_8 \cdot \Re(\omega^8)=0 \]\[ \implies (t_1+t_3-t_5-t_7) \frac{\sqrt{2}}{2} + t_2+t_4 = 0 \implies t_1+t_3-t_5-t_7=0 \]So, $t_1+t_3+t_5+t_7$ is even. We also note that \[\omega \cdot (t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = 0 \implies t_8\omega^1+t_1\omega^2 \cdots +t_7\omega^8\]So, $t_2+t_4+t_6+t_8$ is even by the same logic. Thus, $t_1+t_3+ \cdots +t_8$ is even as desired. $\square$ Thus, if we let $A_n$ denote the number of terms in the expansion of $z_n$, we can set up the following recursion \[ z_{n} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2} = z_{n-1} + \omega^{e_n+4} \cdot z_{n-1} + \omega^{e_n}z_{n-2}\]\[ \implies A_n = 2A_{n-1} + A_{n-2} \]Where $A_0=0$ and $A_1=1$. Note that this recursion implies that for all odd $k$, we have $A_k \equiv A_{k-2} \equiv \cdots \equiv A_1 \equiv 1 \pmod{2}$. Thus, $z_k$ has an odd number of terms in it's $\omega$ expansion. Thus, $z_k \not = 0 = z_0$ for all odd $k$. So, the two ends of the needle cannot swap.

Work in $\mathbb{Z}\left[\frac{\sqrt{2}}{2}\right]^2$. WLOG the needle has length $1$. We will only focus on one endpoint, and prove that it cannot lie on the starting point of the other endpoint. Then note that rotating point $\left(a_1 + b_1\frac{\sqrt{2}}{2}, a_2 + b_2\frac{\sqrt{2}}{2}\right)$ around another results in both $a_1 + b_1$ and $a_2 + b_2$ being invariant modulo $2$. However, since the two endpoints of the needle have distance $1$, the two endpoints have different parities, so they cannot swap.

The answer is no. Main Claim: If $v_1,v_2,\dots v_n$ are vectors with the same magnitude that sum to $0$ and the argument of each vector is a multiple of $45$ degrees, then $n$ is even. Suppose the magnitude is $2$. The idea here is that since $\sqrt{2}$ is irrational, both the integer and $\sqrt{2}$ parts of both $x$ and $y$ coordinates must be zero. Thus $(2,0)$ appears the same number of times as $(-2,0)$, and $(0,2)$ appears the same number of times as $(0,-2)$. Furthermore, $(\sqrt{2},\sqrt{2}),(\sqrt{2},-\sqrt{2})$ in total occur the same number of times as $(-\sqrt{2},\sqrt{2}),(-\sqrt{2},-\sqrt{2})$ in total due to the $x$ coordinate. Thus, the total number of vectors is even. Contract any series of pivots around the same endpoint into a single rotation. Thus, what happens is that the needle repeatedly pivots some multiple of 45 degrees around one endpoint, and then switches the pivot to the other endpoint. The difference between consecutive endpoints is always the same magnitude and has an argument a multiple of $45$ degrees. However, in order for it to get back to its original position with the orientation swapped, the number of such vectors must be odd, contradiction.