I believe this is a counter example. Let $R=\mathbb{C}[x]$, and consider finite-dimensional modules (ie, f.d. vector spaces equipped with a distinguished endomorphism). For convenience, I will identify a module with a matrix, implicitly choosing a basis. Let $$M = \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right], N = [0]$$ be modules of dimension 2 and 1, respectively. Then extensions of $M$ N$by$N$M$ correspond block diagonal matrices of the form $$\left[ \begin{array}{cc} N & C \\ 0 & M \end{array}\right]$$ where $C$ is some $1\times 2$-matrix. Since the automorphisms of $M$ and $N$ act as conjugation by the appropriate matrix, we see that they preserve the rank and nullity of $C$. However
Now, note the two extensions $$C =\left[ 0 \; 0 \right],\;\; C' = \left[ 0 \; 1\right]$$ give isomorphic representations extensions (ie, conjugate matrices), but $C$ and $C'$ have different ranks.
I believe this is a counter example. Let $R=\mathbb{C}[x]$, and consider finite-dimensional modules (ie, f.d. vector spaces equipped with a distinguished endomorphism). For convenience, I will identify a module with a matrix, implicitly choosing a basis. Let $$M = \left[\begin{array}{cc} 0 & 1 \ 0 & 0 \end{array}\right], N = [0]$$ be modules of dimension 2 and 1, respectively. Then extensions of $M$ by $N$ correspond block diagonal matrices of the form $$\left[ \begin{array}{cc} N & C \ 0 & M \end{array}\right]$$ where $C$ is some $1\times 2$-matrix. Since the automorphisms of $M$ and $N$ act as conjugation by the appropriate matrix, we see that they preserve the rank and nullity of $C$. However, the two extensions $$C =\left[ 0 \; 0 \right],\;\; C' = \left[ 0 \; 1\right]$$ give isomorphic representations (ie, conjugate matrices), but $C$ and $C'$ have different ranks.