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No that is not true. Already there are counterexamples on $X=\mathbb{P}^1$. Consider the standard short exact sequence, $$ 0 \to \mathcal{O}(-1) \to \mathcal{O}\oplus \mathcal{O} \to \mathcal{O}(+1) \to 0,$$ and take $H=G=\mathcal{O}(+1)$. Every torsion-free, coherent subsheaf $H'$ of $\mathcal{O}\oplus \mathcal{O}$ is automatically locally free. So your sheaf $H'$ is an invertible sheaf that admits an injective sheaf homomorphism to $\mathcal{O}\oplus \mathcal{O}$. By taking the transpose map, $(H')^\vee$ has nonzero global sections. Thus $(H')^\vee$ has nonnegative degree, i.e., $H'$ has nonpositive degree. Thus the induced morphism $H'\to H$ fails to be an isomorphism already in codimension $1$.