Is conjugation of a positive semi-definite Hermitian matrix equal to conjugation by some rotation? Let $s \in GL_2(\Bbb R)$ be a symmetric positive definite matrix (is this roughly the stretching part of the polar decomposition of some other matrix $x=sk$?). Conjugate $s$ by the reflection $$\gamma=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$ Is there a $k_1 \in K = \left\{\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\right\}$, such that $\gamma s \gamma^{-1}=k_1 s k_1^{-1}$? How do I find it?
More generally, can we answer the question in the title?

Notation: For square matrices $s$ and $t$ of the same order, let $t$-conjugation of $s$ be denoted by $s^t = tst^{-1}$. Then $(s^{t_1})^{t_2} = s^{t_2 t_1}$ [note the reversal of order — this could be remedied by defining conjugation differently, but this answer will be consistent with the definition used in the question].
Analysis of General Case
For a reflection $\gamma$ and a rotation $k$, $s^\gamma = s^k$ if and only if $(s^\gamma)^{k^{-1}} = s^{k^{-1}\gamma} = s$, or $s^\delta = s$, where $\delta = k^{-1}\gamma$, which is another reflection, being the composition of a rotation and a reflection. Now, $s^\delta = s$ is equivalent to $\delta s = s \delta$, so it suffices to find some reflection $\delta$ that commutes with $s$. Then $k = \gamma \delta^{-1}$.
Solution in Two Dimensions
In the two-dimensional special case given in the question, $\gamma = \begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}$, and $s$ is a symmetric positive definite matrix, so let $s = \begin{bmatrix}a & b\\ b & c\end{bmatrix}$ (where $a, c > 0$ and $ac > b^2$, but these will not be used). It is possible to start with a general reflection matrix $\delta$ and apply the commutativity condition obtained earlier to derive possible values of $\delta$. In this answer, I will define a particular $\delta$ and show that it satisfies the condition.
Let $$\cos \theta = \dfrac{a - c}{\sqrt{(a - c)^2 + 4b^2}}, \quad \sin \theta = \dfrac{2b}{\sqrt{(a - c)^2 + 4b^2}}$$
(noting that indeed, $\cos^2 \theta + \sin^2 \theta = 1$).
Let $$\delta = \begin{bmatrix} \cos \theta & \sin \theta\\ \sin \theta & -\cos \theta\end{bmatrix}$$
(noting that $\delta$ is orthogonal and $\det \delta = -1$, which makes it a reflection matrix).
Now
\begin{align*}
s \delta & = \begin{bmatrix}
a \cos \theta + b \sin \theta & -b \cos \theta + a \sin \theta\\
b \cos \theta + c \sin \theta & -c \cos \theta + b \sin \theta
\end{bmatrix}\\
\delta s & = \begin{bmatrix}
a \cos \theta + b \sin \theta & b \cos \theta + c \sin \theta\\
-b \cos \theta + a \sin \theta & -c \cos \theta + b \sin \theta
\end{bmatrix}.
\end{align*}
To verify that $s \delta = \delta s$, it is enough to check whether $-b \cos \theta + a \sin \theta = b \cos \theta + c \sin \theta$ (since both matrices have the same diagonal, and the $(1,2)$ elements are equal if and only if the $(2,1)$ elements are equal).
Rearranging this equation, we get $(a - c) \sin \theta = 2b \cos \theta$, which does hold since $\tan \theta = \dfrac{2b}{a - c}$ (by our definition of $\cos \theta$ and $\sin \theta$). Thus, $\delta$ is the required matrix, as claimed.
Now, $k = \gamma \delta^{-1}$, where $\delta^{-1} = \delta^T = \delta$, so we have
\begin{equation*}
k = \begin{bmatrix}
\cos \theta & \sin \theta\\
- \sin \theta & \cos \theta
\end{bmatrix}.
\end{equation*}
