Let's consider the following differential equation $$ \frac{dA}{dt} + A \cdot X = 0, $$ where $A(t)$ and $X(t)$ are $N \times N$ matrices.
I'm looking for a $A \rightarrow B$ substitution which transforms this equation to the form $$ \frac{dB}{dt} + X \cdot B = 0. $$
One's first intuition is to use $A = B^{-1}$, then we have $$ -B^{-1} \cdot \frac{dB}{dt} \cdot B^{-1} + B^{-1} \cdot X = 0. $$ Multiplying the last equality by $B$ from both sides we obtain $$ \frac{dB}{dt} - X \cdot B = 0. $$ But in the final expression the sign at $X \cdot B$ is changed.
Can one propose an appropriate transformation or show that it doesn't exist?
Consider the case $$X = \pmatrix{ 1/(t+1) & 1\cr 0 & 0\cr}$$ The solution of $\dfrac{dA}{dt} = X(t) A(t)$ with $A(0) = I$ is $$ A(t) = \pmatrix{(1+t)\ln(1+t) & 1+t\cr 1 & 0\cr}$$ The solution of $\dfrac{dB}{dt} = B(t) X(t)$ with $B(0)=I$ is $$ B(t) = \pmatrix{t + 1 & t^2/2 + t\cr 0 & 1\cr}$$
There is no nice transformation that takes $A(t)$ to $B(t)$ or vice versa. In particular, the entries of $B(t)$ are polynomials while those of $A(t)$ are not all analytic.
EDIT: Similarly, if you take $X_{12} = e^t$ instead of $1$, you get an example where all entries of $B(t)$ are elementary functions, but one of the entries of $A$ is not.