exponential commutes integral Let $B(t)$ be a $n\times n$ function matrix with respect to $t$, consider $B(t)$ is differentiable , the following conclusion is obvious 
$$\text{if}\ \ B'B=BB',\ \ \text{then} \ \ B'e^B=e^BB'$$
So my question is if the converse direction correct? i.e. can be found some $B$ such that
$$B'B\neq BB'\ \ \text{but}\ \ B'e^B=e^BB'.$$.
In deed, I found a complex matrix satisfy this condition
$$B=\begin{pmatrix} \frac{2i\pi}{t-1} & \frac{2i\pi}{t-1}\\
0 & \frac{2i\pi t}{t-1}\end{pmatrix}$$
However, I am wondering if there exist real matrix (I mean the coefficients are real ) example???

Updates
Above is in differential form, now I consider the integral form:
Can we find some $B$ such that:
$$B\int_{t_0}^tB(s)ds\neq \int_{t_0}^tB(s)dsB\;\;but\;\;Be^{\int_{t_0}^tB(s)ds}=e^{\int_{t_0}^tB(s)ds}B$$
@John Ma, I tried you example, but failed to the integral case.
 A: I only give a $3\times 3$ example. Let 
$$R(t) = \begin{bmatrix} 0 & \cos t & \sin t \\ -\cos t & 0 & 0 \\ -\sin t & 0& 0 \end{bmatrix}\Rightarrow R(t) = \begin{bmatrix} 0 & -\sin t & \cos t \\ \sin t & 0 & 0 \\ -\cos t & 0& 0 \end{bmatrix}$$
One can check that $R' R \neq RR'$ (check the $(2, 3)$ entry). Now we show that $e^{ B(t)} = I$ for all $t\in I$, where $B(t) = 2\pi R(t)$. Fix $t$, let 
$$v_1 = \begin{bmatrix} 0 \\ \cos t \\ \sin t \end{bmatrix} ,\ \ v_2 = \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix},\ \ v_3 = \begin{bmatrix} 0 \\ -\sin t \\ \cos t \end{bmatrix}$$
Then by a direct checking, 
$$B(t) v_1 = 2\pi v_2, \ \ B(t) v_2 = -2\pi v_1, \ \ B(t) v_3 = 0.$$
Thus there is an invertible matrix $A = A(t)$ and 
$$K = \begin{bmatrix} 0 & -2\pi & 0 \\ 2\pi & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
so that 
$$B(t) = A K A^{-1}\Rightarrow e^{B(t)} = A e^{K} A^{-1} = I$$
as $e^K = I$. Thus $BB' \neq B'B$ for all $t$ and $B' e^B = e^B B'$ for all $t$. 
Remark The above construction look miraculous, but that actually follows from a special property concerning the rotational group $SO(3)$ on $\mathbb R^3$. Indeed, we can take any $R(t)$ of the form: 
$$R(t) = \begin{bmatrix} 0 & a(t) & b(t) \\ -a(t) & 0 & c(t) \\ -b(t) & -c(t) & 0 \end{bmatrix}$$
so that $a^2 + b^2 + c^2 = 1$ and $(a' , b' , c') \neq (0,0,0)$. Then again set $B(t) = 2\pi R(t)$. The fact that $B' B \neq B B'$ is actually related to the cross products on $\mathbb R^3$. 
