$\pi/2$ rotation of spin $1/2$ $\def\ket#1{{\lvert #1 \rangle}}$
In John S. Townsend's A Modern Approach to quantum Mechanics, an operator is introduced in Ch. 2 that rotates the spin by $\pi/2$ in the $x$-$z$ plane.  These examples are given:
$$\ket{+x} = \hat{R}(\tfrac{\pi}{2}\mathbf{j}) \ket{+z}$$
$$\ket{-x} = \hat{R}(\tfrac{\pi}{2}\mathbf{j}) \ket{-z}$$
My question is: should that same operator also rotate:

*

*$\ket{+x}$ into $\ket{-z}$ and

*$\ket{-x}$ into $\ket{+z}$
so that applying it four times would get you back to the same state? If not? Why not? If so, what am I doing wrong here:
$$
\hat{R}(\tfrac{\pi}{2}\mathbf{j})\ket{+x} =
\hat{R}(\tfrac{\pi}{2}\mathbf{j})\left[\frac{1}{\sqrt{2}}\left(\ket{+z} + \ket{-z} \right) \right] \\
= \frac{1}{\sqrt{2}}\left(\ket{+x} + \ket{-x} \right) \\
= \ket{+z}
$$
and similarly for $\hat{R}(\tfrac{\pi}{2}\mathbf{j})\ket{-x} = \ket{-z}$.
It looks like applying this operator twice brings you back to the same state.  I'm a bit confused.
 A: $\newcommand{\ket}[1]{\lvert #1 \rangle}$

In John S. Townsend's A Modern Approach to quantum Mechanics, an operator is introduced in Ch. 2 that rotates the spin by $\pi/2$ in the $x$-$z$ plane.  These examples are given:
$$\ket{+x} = \hat{R}(\tfrac{\pi}{2}\mathbf{j}) \ket{+z}$$
$$\ket{-x} = \hat{R}(\tfrac{\pi}{2}\mathbf{j}) \ket{-z}$$

The last equation is only true up to a phase (a factor of -1 in this case).

My question is: should that same operator also rotate:

*

*$\ket{+x}$ into $\ket{-z}$ and

*$\ket{-x}$ into $\ket{+z}$

Yes, up to a phase.

so that applying it four times would get you back to the same state? If not? Why not?

Yes, up to a phase.
For example, for spin-1: ${R(\frac{\pi}{2}\mathbf{j})}^4 = R(2\pi) = 1$
But, for example, for spin-1/2: ${R(\frac{\pi}{2}\mathbf{j})}^4 = R(2\pi) = -1$, which is funny, but nevertheless true.

...what am I doing wrong here...

The phase was not properly accounted for. For example:
$$
R(\frac{\pi}{2}\mathbf{j})\ket{+z} = \ket{+x}
$$
$$
R(\frac{\pi}{2}\mathbf{j})\ket{-z} = -\ket{-x}
$$
You can see this from, for example, the explicit matrix for the rotation operator:
$$
R(\frac{\pi}{2}\mathbf{j}) = 
\left(\begin{matrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}\right) 
$$

Update:
In the $(|+z>, |-z>)$ basis the rotation matrix OP is interested in is
$$
R= \left(\begin{matrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}\right)
$$
Therefore:
$$
R\ket{+z} = \left(\begin{matrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}\right)
\left(\begin{matrix}1 \\ 0 \end{matrix}\right)
=
\left(\begin{matrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{matrix}\right)
\equiv \ket{+x}
$$
Therefore:
$$
R\ket{-z} = \left(\begin{matrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}\right)
\left(\begin{matrix}0 \\ 1 \end{matrix}\right)
=
\left(\begin{matrix} -1/\sqrt{2} \\ 1/\sqrt{2} \end{matrix}\right)
=-\left(\begin{matrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{matrix}\right)
\equiv -\ket{-x}
$$

Note also, the rotation matrix for any angle in any direction, for spin-1/2, can be written as:
$$
R(\theta\mathbf{n}) = \cos(\theta/2)\left(\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}\right) - i\sin(\theta/2)\mathbf{\sigma}\cdot \mathbf{n}\;,
$$
where $\mathbf{\sigma}$ is a vector of the Pauli matrices.
