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.