It's known that $SO(3)$ is isomorphic to SU(2), I am trying to establish a relation from exponentiation of their Lie Algebra, the formula I would like to prove is:

$$e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2}=R_{ij}(\hat{n},\theta)\sigma_{i} $$
Here is the prerequisite derivation, for $SO(3)$:
The $SO(3)$ matrix in adjoint representation is given by:
\begin{equation}
    (T_{i})_{jk}=-i\epsilon_{ijk}
\end{equation}
notice it satisfy the commutation relation:
\begin{equation}
    [T_{i},T_{j}]_{\alpha \beta}=(T_{i})_{\alpha \gamma}(T_{j})_{\gamma \beta}-(T_{j})_{\alpha \gamma}(T_{i})_{\gamma \beta}=-\epsilon_{i\alpha \gamma}\epsilon_{j\gamma \beta}+\epsilon_{j\alpha \gamma}\epsilon_{i\gamma \beta}
\end{equation}
using the generalise Kroneckar delta:
\begin{equation}
    [T_{i},T_{j}]_{\alpha \beta}=\epsilon_{\gamma i\alpha}\epsilon_{\gamma j \beta}-\epsilon_{\gamma j \alpha}\epsilon_{\gamma i \beta}=\delta_{j\beta}\delta_{i\alpha}-\delta_{i\beta}\delta_{j\beta}=i\epsilon_{ijk}(T_{k})_{\alpha \beta}
\end{equation} 
given that $n$ is a unit vector, notice the square satisfy:
\begin{equation}
    (n\cdot T)_{jm}^{2}=(n_{i}T_{i})_{jk}(n_{x}T_{x})_{kl}=-n_{i}n_{x}\epsilon_{kij}\epsilon_{klx}=\delta_{jl}-n_{l}n_{j}
\end{equation}
\begin{equation}
    (n\cdot T)_{jm}^{3}=(n_{i}T_{i})_{jk}(n_{x}T_{x})_{kl}(n_{y}T_{y})_{lm}=n_{i}n_{x}n_{y}(-i^{3})\epsilon_{kij}\epsilon_{klx}\epsilon_{lmy}
\end{equation}
using the Kroneckar delta:
\begin{equation}
     (n\cdot T)_{jm}^{3}=in_{i}n_{x}n_{y}\epsilon_{lmy}(\delta_{il}\delta_{jx}-\delta_{ix}\delta_{jl})=-in_{x}^{2}n_{y}\epsilon_{jmy}=|n|^{2}(n\cdot T)_{jm}
\end{equation}
the eigenvalue of $(n\cdot T)$ then equivalent to:
\begin{equation}
    (n\cdot T)[(n\cdot T)^{2}-1]=0 \;\;,\;\; n\cdot T=-1,0,1\;\;,\;\; J=1
\end{equation}
using exponential to obtain the rotation:
\begin{equation}
    R_{ij}(\hat{n},\theta);=\exp(-i\theta \hat{n}\cdot T)_{ij}=\sum_{n=0}^{\infty}\frac{(-i\theta \hat{n}\cdot T)_{ij}^{n}}{n!}
\end{equation}
expanding few terms to observe and grouping like terms:
\begin{equation}
    I_{ij}-(\hat{n}\cdot T)_{ij}^{2}+(\hat{n}\cdot T)_{ij}^{2}(1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}+\dots)-i(\hat{n}\cdot T)_{ij}(\theta-\frac{\theta^{3}}{3!}+\dots)
\end{equation}
therefore the rotation is:
\begin{equation}
    R_{ij}(\hat{n},\theta)=[1-(\hat{n}\cdot T)^{2}]_{ij}+(\hat{n}\cdot T)_{ij}^{2}\cos\theta-i(\hat{n}\cdot T)_{ij}\sin\theta
\end{equation}
further simplify in compact form:
\begin{equation}
       R_{ij}(\hat{n},\theta)=[\delta_{ij}-(\delta_{ij}-n_{i}n_{j})]+(\delta_{ij}-n_{i}n_{j})\cos\theta-i(-i\epsilon_{kij}n_{k})\sin\theta
\end{equation}
which yield the Rodrigues Formula:
\begin{equation}
    R_{ij}(\hat{n},\theta)=n_{i}n_{j}+(\delta_{ij}-n_{i}n_{j})\cos\theta-\epsilon_{ijk}n_{k}\sin\theta
\end{equation}
now also I obtained the $SU(2)$ The expression:
\begin{equation}
e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2}
\end{equation}
using the BCH formula:
\begin{equation}
\sigma_{j}+(-\frac{i\theta n_{i}}{2})[\sigma_{i},\sigma_{j}]+\frac{(-\frac{i\theta}{2}n_{i})^{2}}{2!}[\sigma_{i},[\sigma_{i},\sigma_{j}]]+\frac{((-\frac{i\theta n_{i}}{2}))^{3}}{3!}[\sigma_{i},[\sigma_{i},[\sigma_{i},\sigma_{j}]]]+\dots
\end{equation}
corresponding commutators are
\begin{equation}
    [\sigma_{i},\sigma_{j}]=i\epsilon_{ijk}\sigma_{k} \;\;,\;\;  i\epsilon_{ijk} [\sigma_{i},\sigma_{k}]=-\epsilon_{ijk}\epsilon_{ikj}\sigma_{j}=\sigma_{j}
\end{equation}
observing the pattern and grouping like terms
\begin{equation}
  \sigma_{j}[1-\frac{1}{2!}(\frac{\theta}{2})^{2}+\dots]+i\epsilon_{ijk}\sigma_{k}n_{i}(-i)[\frac{1}{1!}(\frac{\theta}{2})-\frac{1}{3!}(\frac{\theta}{2})^{3}+\dots]
\end{equation}
using the Taylor expansion of sine and cosine:
\begin{equation}
    e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2}=\sigma_{j}\cos\theta+\epsilon_{ijk}n_{i}\sigma_{k}\sin\theta
\end{equation}

Nonetheless, I am not able to prove or disprove the first formula.