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Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.

Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.

A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9

Update on part 1:

I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'. I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.

ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.

Part 3: Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\\#E(\mathbb F_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would means that it comes from some multiplication by $n$ map and hence that the index should be a square.

In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.

Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.

Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.

A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9

Update on part 1:

I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'. I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.

ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.

Part 3: Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\\#E(\mathbb F_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would mean that it comes from some multiplication by $n$ map and hence that the index should not be squarefree.

In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.

Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.

Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.

A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9

Update:

I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'. I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.

Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.

Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.

A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9

Update on part 1:

I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'. I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.

ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.

Part 3: Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\\#E(\mathbb F_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would means that it comes from some multiplication by $n$ map and hence that the index should be a square.

In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.

Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.

Now let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.

A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9

Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.

Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.

A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9

Update:

I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'. I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.