Similarity transformation of Lorentz transformation

Obtain The Lorentz Transformation in which the velocity is at an infinitesimal angle $$d\theta$$ counter clockwise from the $$x$$ axis, by means of a similarity transformation applied to $$L=\begin{bmatrix}\gamma & -\gamma \beta & 0& 0 \\ -\gamma \beta & \gamma & 0 &0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \tag{1}$$

show directly that the resulting matrix is orthogonal and that the inverse matrix is obtained by substituting $$-v$$ for $$v$$

So I thought this was a pretty straight forward question, Since the vector is counter clockwise from the $$x$$ axis I thought I would first apply a counterclockwise infinitesimal rotation about the $$z$$ axis to match the $$x$$ axis with the vector then apply $$L$$ and then rotate the resulting transformation clockwise round the $$z$$ axis to get back to the original basis.

\begin{align}T =RLR^{-1} &= \begin{bmatrix}1&0&0&0 \\ 0&1&-d \theta &0 \\ 0&d \theta&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix}\gamma & -\gamma \beta & 0& 0 \\ -\gamma \beta & \gamma & 0 &0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}\begin{bmatrix}1&0&0&0 \\ 0&1&d \theta &0 \\ 0& -d \theta&1&0 \\ 0&0&0&1 \end{bmatrix} \\ \\&=\begin{bmatrix}\gamma&-\gamma \beta& -\gamma\beta d\theta &0 \\ -\gamma \beta & (\gamma+d \theta)&(\gamma d \theta - d \theta)&0 \\-\gamma \beta d \theta& (\gamma d \theta -d \theta)& 1& 0 \\ 0&0&0&1\end{bmatrix} \end{align} Where I have ignored second order terms of $$d \theta$$. However, this $$T$$ is not orthogonal ($$T^{T}T \ne I$$). Furthermore, replacing $$v$$ with $$-v$$ I got the following

\begin{align}M=T(v)T(-v) &= \begin{bmatrix}\gamma&-\gamma \beta& -\gamma\beta d\theta &0 \\ -\gamma \beta & (\gamma+d \theta)&(\gamma d \theta - d \theta)&0 \\-\gamma \beta d \theta& (\gamma d \theta -d \theta)& 1& 0 \\ 0&0&0&1\end{bmatrix} \begin{bmatrix}\gamma&\gamma \beta& \gamma\beta d\theta &0 \\ \gamma \beta & (\gamma+d \theta)&(\gamma d \theta - d \theta)&0 \\\gamma \beta d \theta& (\gamma d \theta -d \theta)& 1& 0 \\ 0&0&0&1\end{bmatrix} \\ \\ &= \begin{bmatrix} 1& -\gamma \beta d \theta &0&0 \\ \gamma \beta d \theta&1&0&0\\0&0&1&0 \\ 0&0&0&1\end{bmatrix} \end{align}

Which doesnt give me the inverse i.e $$T(v)T(-v) \ne I$$ therefore $$T(-v)\ne T(v)^{-1}$$. However, the matrix $$M$$ does look similar to a Thomas Precession. So I am a bit confused; I'm not sure where I went wrong in calculating the similarity transformation $$T$$.

The reason your final T is not orthogonal is that your original L is not orthogonal ($$L^T=L \neq L^{-1}$$). In fact Lorentz boosts of $$(ct,x)$$ really are not orthogonal, and the matrix you have written for L is correct. If L were orthogonal it would leave $$t^2+x^2$$ invariant. Instead L leaves $$t^2-x^2$$ invariant.
The request to prove orthogonality in your question doesn't go with the real matrix L that you wrote. Instead orthogonality goes with $$L_{Historical}$$ rotating by an imaginary angle.
$$\begin{bmatrix} x'\\ ict'\\ \end{bmatrix} =L_{Historical}\begin{bmatrix} x\\ ict\\ \end{bmatrix} = \begin{bmatrix} cos(i\lambda) & -sin(i\lambda)\\ sin(i\lambda) & cos(i\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix} = \begin{bmatrix} cosh(\lambda) & -i\ sinh(\lambda)\\ i\ sinh(\lambda) & cosh(\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix}$$ $$\begin{bmatrix} x'\\ ict'\\ \end{bmatrix} = \begin{bmatrix} \gamma & -i\beta\gamma\\ i\beta\gamma & \gamma\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix}$$