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The answer is trivially no for abelian groups. An infinite series of counterexamples can be found by taking $G=\mathbb{Z}_4 \times (\mathbb{Z}_2)^n,$ with $$H_1 = \mathbb{Z}_4 \times (\mathbb{Z}_2)^{n-1}, \quad H_2 = \mathbb{Z}_2 \times (\mathbb{Z}_2)^{n}.$$

Since you put "abelian" into parentheses, let me show that the answer is no also for nonabelian groups. Consider the finite group of order $16$ $$G := \langle a,b,x \mid a^4 = b^2 = x^2 = 1, \, ab = ba, \, bx = xb, \, xax^{-1} = ab \rangle,$$ whose GAP label is $G(16, \, 3)$, and take the two normal subgroups $$H_1 = \langle a, \, b \rangle \simeq \mathbb{Z}_4 \times \mathbb{Z}_2, \quad H_2 = \langle a^2, \, b, \, x \rangle \simeq \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2.$$ They are not isomorphic as abstract groups, so no automorphism of $G$ can send $H_1$ to $H_2$. However, we have $$G/H_1 \simeq G/H_2 \simeq \mathbb{Z}_2.$$

Let me give another non-abelian counterexample in which $H_1$ and $H_2$ are isomorphic as abstract groups. Consider the group of order $16$ $$G := \langle a,\,x \mid a^4 = x^4 = 1, \, xax^{-1} = a^{-1} \rangle,$$ namely, the non-trivial semidirect product of $\mathbb{Z}_4$ with itself, whose GAP label is $G(16, \, 4)$. Now, take the two subgroups $$H_1= \langle a^2, \, x \rangle, \quad H_2= \langle a, \, x^2 \rangle.$$ They are both isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$, so they are normal and, again, the quotients are both isomorphic to $\mathbb{Z}_2$. However, $H_2$ is characteristic, so every automorphism of $G$ sends $H_2$ to itself.

The answer is rather obviously no for abelian groups. An infinite series of counterexamples can be found by taking $G=\mathbb{Z}_4 \times (\mathbb{Z}_2)^n,$ with $$H_1 = \mathbb{Z}_4 \times (\mathbb{Z}_2)^{n-1}, \quad H_2 = \mathbb{Z}_2 \times (\mathbb{Z}_2)^{n}.$$

Since you put "abelian" into parentheses, let me show that the answer is no also for non-abelian groups, providing a counterexample in which $H_1$ and $H_2$ are isomorphic as abstract groups. Consider the group of order $16$ $$G := \langle a,\,x \mid a^4 = x^4 = 1, \, xax^{-1} = a^{-1} \rangle,$$ namely, the non-trivial semidirect product of $\mathbb{Z}_4$ with itself, whose GAP label is $G(16, \, 4)$. Now, take the two subgroups $$H_1= \langle a^2, \, x \rangle, \quad H_2= \langle a, \, x^2 \rangle.$$ They are both isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$, so they are normal and the quotients are both isomorphic to $\mathbb{Z}_2$. However, $H_2$ is characteristic, so every automorphism of $G$ sends $H_2$ to itself.

The answer is no, also for finite groups. Consider the finite group of order $16$ $$G := \langle a,b,x \mid a^4 = b^2 = x^2 = 1, \, ab = ba, \, bx = xb, \, xax^{-1} = ab \rangle,$$ whose GAP label is $G(16, \, 3)$, and take the two normal subgroups $$H_1 = \langle a, \, b \rangle \simeq \mathbb{Z}_4 \times \mathbb{Z}_2, \quad H_2 = \langle a^2, \, b, \, x \rangle \simeq \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2.$$ They are not isomorphic as abstract groups, so no automorphism of $G$ can send $H_1$ to $H_2$. However, we have $$G/H_1 \simeq G/H_2 \simeq \mathbb{Z}_2.$$

Let me give another non-abelian counterexample in which $H_1$ and $H_2$ are isomorphic as abstract groups. Consider the group of order $16$ $$G := \langle a,\,x \mid a^4 = x^4 = 1, \, xax^{-1} = a^{-1} \rangle,$$ namely, the non-trivial semidirect product of $\mathbb{Z}_4$ with itself, whose GAP label is $G(16, \, 4)$. Now, take the two subgroups $$H_1= \langle a^2, \, x \rangle, \quad H_2= \langle a, \, x^2 \rangle.$$ They are both isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$, so they are normal and, again, the quotients are both isomorphic to $\mathbb{Z}_2$. However, $H_2$ is characteristic, so every automorphism of $G$ sends $H_2$ to itself.

The answer is trivially no for abelian groups. An infinite series of counterexamples can be found by taking $G=\mathbb{Z}_4 \times (\mathbb{Z}_2)^n,$ with $$H_1 = \mathbb{Z}_4 \times (\mathbb{Z}_2)^{n-1}, \quad H_2 = \mathbb{Z}_2 \times (\mathbb{Z}_2)^{n}.$$

Since you put "abelian" into parentheses, let me show that the answer is no also for nonabelian groups. Consider the finite group of order $16$ $$G := \langle a,b,x \mid a^4 = b^2 = x^2 = 1, \, ab = ba, \, bx = xb, \, xax^{-1} = ab \rangle,$$ whose GAP label is $G(16, \, 3)$, and take the two normal subgroups $$H_1 = \langle a, \, b \rangle \simeq \mathbb{Z}_4 \times \mathbb{Z}_2, \quad H_2 = \langle a^2, \, b, \, x \rangle \simeq \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2.$$ They are not isomorphic as abstract groups, so no automorphism of $G$ can send $H_1$ to $H_2$. However, we have $$G/H_1 \simeq G/H_2 \simeq \mathbb{Z}_2.$$

Let me give another non-abelian counterexample in which $H_1$ and $H_2$ are isomorphic as abstract groups. Consider the group of order $16$ $$G := \langle a,\,x \mid a^4 = x^4 = 1, \, xax^{-1} = a^{-1} \rangle,$$ namely, the non-trivial semidirect product of $\mathbb{Z}_4$ with itself, whose GAP label is $G(16, \, 4)$. Now, take the two subgroups $$H_1= \langle a^2, \, x \rangle, \quad H_2= \langle a, \, x^2 \rangle.$$ They are both isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$, so they are normal and, again, the quotients are both isomorphic to $\mathbb{Z}_2$. However, $H_2$ is characteristic, so every automorphism of $G$ sends $H_2$ to itself.

The answer is no, also for finite groups. Consider the finite group of order $16$ $$G := \langle a,b,x \mid a^4 = b^2 = x^2 = 1, \, ab = ba, \, bx = xb, \, xax^{-1} = ab \rangle,$$ whose GAP label is $G(16, \, 3)$, and take the two normal subgroups $$H_1 = \langle a, \, b \rangle \simeq \mathbb{Z}_4 \times \mathbb{Z}_2, \quad H_2 = \langle a^2, \, b, \, x \rangle \simeq \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2.$$ They are not isomorphic as abstract groups, so no automorphism of $G$ can send $H_1$ to $H_2$. However, we have $$G/H_1 \simeq G/H_2 \simeq \mathbb{Z}_2.$$

Let me give another non-abelian counterexample in which $H_1$ and $H_2$ are isomorphic as abstract groups. Consider the group of order $16$ $$G := \langle a,\,x \mid a^4 = x^4 = 1, \, xax^{-1} = a^{-1} \rangle,$$ namely, the non-trivial semidirect product of $\mathbb{Z}_4$ with itself, whose GAP label is $G(16, \, 4)$. Now, take the two subgroups $$H_1= \langle a^2, \, x \rangle, \quad H_2= \langle a, \, x^2 \rangle.$$ They are both isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$, so they are normal and, again, the quotients are both isomorphic to $\mathbb{Z}_2$. However, $H_2$ is characteristic, so no automorphism of $G$ can send $H_1$ to $H_2$.

The answer is no, also for finite groups. Consider the finite group of order $16$ $$G := \langle a,b,x \mid a^4 = b^2 = x^2 = 1, \, ab = ba, \, bx = xb, \, xax^{-1} = ab \rangle,$$ whose GAP label is $G(16, \, 3)$, and take the two normal subgroups $$H_1 = \langle a, \, b \rangle \simeq \mathbb{Z}_4 \times \mathbb{Z}_2, \quad H_2 = \langle a^2, \, b, \, x \rangle \simeq \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2.$$ They are not isomorphic as abstract groups, so no automorphism of $G$ can send $H_1$ to $H_2$. However, we have $$G/H_1 \simeq G/H_2 \simeq \mathbb{Z}_2.$$

Let me give another non-abelian counterexample in which $H_1$ and $H_2$ are isomorphic as abstract groups. Consider the group of order $16$ $$G := \langle a,\,x \mid a^4 = x^4 = 1, \, xax^{-1} = a^{-1} \rangle,$$ namely, the non-trivial semidirect product of $\mathbb{Z}_4$ with itself, whose GAP label is $G(16, \, 4)$. Now, take the two subgroups $$H_1= \langle a^2, \, x \rangle, \quad H_2= \langle a, \, x^2 \rangle.$$ They are both isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$, so they are normal and, again, the quotients are both isomorphic to $\mathbb{Z}_2$. However, $H_2$ is characteristic, so every automorphism of $G$ sends $H_2$ to itself.