Does $\text{SO}_2(\mathbb{Q}_5)$ contain non-trivial elements? I was trying to find an element of $\text{SO}_2(\mathbb{Q}_5)$ for the $5$-adic numbers.  By analogy with $\text{SO}_2(\mathbb{R})$ 
$$ \left[ \begin{array}{rr} a & -b \\ b & a \end{array} \right] \text{ with } a^2 + b^2 = 1$$
When we solve this equation modulo $5$, the perfect squares are $\square = \{ 0,1,4\}$ and the only solutions are $0^2 + (\pm 1)^2 = 1$ up to permutations.
Momentarily, I considered $4^2 + (\sqrt{-15})^2 = 1$ but we have $\sqrt{-15} \notin \mathbb{Q}_5$ or else we'd have the valuation $|\sqrt{-15}|_5 = \frac{1}{\sqrt{5}}$.
Certainly there are trivial elements such as the identity element and the $90^\circ$ rotation:
$$ \left[ \begin{array}{rr} 0 & \mp 1 \\ \pm 1 & 0 \end{array} \right] , 
\left[ \begin{array}{rr} \pm 1 & 0 \\ 0 & \pm 1 \end{array} \right] \in \text{SO}_2(\mathbb{Q}_5) $$
By the looks of it $\text{SO}_2(\mathbb{Q}_7)$ only has trivial elements as well as the perfect squares are $\square_7 = \{0,1,2,4\}$.

EDIT As the answer points out $\text{SO}(\mathbb{Q}) \subset \text{SO}(\mathbb{Q}_5)$.  Can we find elements of $\text{SO}(\mathbb{Q}_5) \backslash \text{SO}(\mathbb{Q})$ ?  
I think the answer work because $(5,13)=1$ and so $\frac{1}{13} \in \mathbb{Z}_5$.

$$\text{SO}_2(\mathbb{Q}_5) = \{ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} | (a,b) \in \mathbb{Q}_5, a^2 + b^2 = 1 \}$$
Let $$f(x) = (1+x)^{1/2} = \sum_{k=0}^\infty {1/2 \choose k} x^k, \qquad |x|_5 < 1$$
Then $|{1/2 \choose k}|_5 = |\prod_{m=0}^{k-1} \frac{1/2-m}{k-m}|_5= |\prod_{m=0}^{k-1} \frac{\frac{5^r+1}{2}+O(5^{r+1})-m}{k-m}|_5 \le 1$ whence $f(x)$ converges for $|x|_5 < 1$ and since $\mathbb{Q}_5,|.|_5$ is complete then $x \in 5 \mathbb{Z}_5 \implies (1+x)^{1/2} \in \mathbb{Q}_5$.
Whence 


*

*if $|a|_5 > 1$ then $a^2 = 5^{-2m} d^2(1+x)$ and $b = \pm 5^{-m} d (1+x)^{1/2}$ 

*if $|a|_5 < 1$ then $a^2-1 = 2^2 (1-a^2/4)$ and $b= \pm 2 (1-a^2/4)^{1/2}$

*if $|a|_5 = 0$, iff $a^2-1 \equiv d^2 \bmod 5$ then $a^2 -1 = d^2 (1+x)$ and $b = \pm d (1+x)^{1/2}$
