# How to prove expression for relative pose in one frame given relative pose in another frame

Given the poseo of point 1 between $$t_k$$ and $$t_{k+1}$$ $$\begin{equation} \mathbf{P}_{1}\left(t_{k}, t_{k+1}\right)=\left[\begin{array}{cc} \mathbf{R}_{1}\left(t_{k}, t_{k+1}\right) & \mathbf{T}_{1}\left(t_{k}, t_{k+1}\right) \\ 0 & 1 \end{array}\right] . \end{equation}$$ and the relative pose between point 1 and 2, $$\mathbf{P_{12}}$$ why is the relative pose between $$t_k$$ and $$t_{k+1}$$ in point 2 given as:

$$\begin{equation} \mathbf{P}_{2}\left(t_{k}, t_{k+1}\right)=\mathbf{P}_{12} \mathbf{P}_{1}\left(t_{k}, t_{k+1}\right) \mathbf{P}_{12}^{-1} \end{equation}$$.

Why is this expression true?

My attempt: I have tried multiplying moth sides by $$\mathbf{P}_{12}^{-1} = \mathbf{P}_{21}$$.

I get:

$$\mathbf{P}_{21}\mathbf{P}_{2}\left(t_{k}, t_{k+1}\right)$$ which in my opinion equals $$\mathbf{P}_{1}\left(t_{k}, t_{k+1}\right)$$ but instead according to the expression it equals $$\mathbf{P}_{1}\left(t_{k}, t_{k+1}\right) \mathbf{P}_{12}^{-1}$$. Starting from the point $$2k+1$$, we can count the transformations anticlockwise, ending up again at the same point; hence, the aggregate transformation shall be the identity matrix $$I$$.
$$P_{12}^{-1} \cdot P_1^{-1} \cdot P_{12} \cdot P_2 = I,$$