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 $N$ by $M$ 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 $C$.
Now, note the two extensions $$ C =\left[ 0 \; 0 \right],\;\; C' = \left[ 0 \; 1\right] $$ give isomorphic extensions (ie, conjugate matrices), but $C$ and $C'$ have different ranks.