Statement: Let $C$ be union of 3 smooth conics in $\mathbb{P}^2$ meeting each other in nodes. Then there exists a line $L\cong \mathbb{P}^1$ such that the configuration of 6 points $x_1,…,x_6\in L$ obtained by intersecting $L$ with $C$ has an automorphism of order 5.

The above fact is surprising (to me at least) since by a dimension count, one cannot get an arbitrary configuration of six points and the configuration of points with an order 5 automorphism is unique. The fact seems to follow from some work a student of mine is doing involving genus 2 K3 surfaces (more below), but the statement seems so classical, that it feels like there should be a direct elementary proof.

Question: Is there a "classical" proof of the above statement?

In fact, the statement seems to hold for any (possibly reducible) sextic curve $C$ with singularities no worse than nodes as long as (1) there is at least one node, and (2) there are no components which are lines.

Background:

Given a sextic plane curve $C$ with at worst nodal singularities, we get a K3 surface by taking the double cover of the plane branched over $C$ and then resolving the ordinary double point singularities. The double covers of the lines in the plane give rise to a two dimensional linear system of genus 2 curves on the K3 which give a rational map $\mathbb{P}^2--\to \overline{M}_2$. One can compute the class of the closure of the image (call it $D$) in the chow ring of $\overline{M}_2$ by intersecting it with test curves in $\overline{M}_2$ (this is tricky so it is possible that we are making mistakes here) and one gets a formula for $D$ as a linear combination of boundary divisors. If $C$ is smooth, the linear combination has integer coefficients, but under the assumption that there is at least one node and no lines in $C$, the coefficients have 5's in the denominator. I think the only way that can happen is if $D$ meets the (unique I think) genus 2 curve with an order 5 automorphism. This gives rise to the statement about the points on the line in $\mathbb{P}^2$.

Question 2: Does the statement hold when $C$ contains lines? For example if $C$ is the union of 6 disjoint lines?

The above proof breaks down in this case.

Edit: In light of Jason's answer, the below asserted statement must be wrong, which means that the intersection theory argument alluded to in the background section must be wrong. What follows is the question as originally stated:

Statement: Let $C$ be union of 3 smooth conics in $\mathbb{P}^2$ meeting each other in nodes. Then there exists a line $L\cong \mathbb{P}^1$ such that the configuration of 6 points $x_1,…,x_6\in L$ obtained by intersecting $L$ with $C$ has an automorphism of order 5.

The above fact is surprising (to me at least) since by a dimension count, one cannot get an arbitrary configuration of six points and the configuration of points with an order 5 automorphism is unique. The fact seems to follow from some work a student of mine is doing involving genus 2 K3 surfaces (more below), but the statement seems so classical, that it feels like there should be a direct elementary proof.

Question: Is there a "classical" proof of the above statement?

In fact, the statement seems to hold for any (possibly reducible) sextic curve $C$ with singularities no worse than nodes as long as (1) there is at least one node, and (2) there are no components which are lines.

Background:

Given a sextic plane curve $C$ with at worst nodal singularities, we get a K3 surface by taking the double cover of the plane branched over $C$ and then resolving the ordinary double point singularities. The double covers of the lines in the plane give rise to a two dimensional linear system of genus 2 curves on the K3 which give a rational map $\mathbb{P}^2--\to \overline{M}_2$. One can compute the class of the closure of the image (call it $D$) in the chow ring of $\overline{M}_2$ by intersecting it with test curves in $\overline{M}_2$ (this is tricky so it is possible that we are making mistakes here) and one gets a formula for $D$ as a linear combination of boundary divisors. If $C$ is smooth, the linear combination has integer coefficients, but under the assumption that there is at least one node and no lines in $C$, the coefficients have 5's in the denominator. I think the only way that can happen is if $D$ meets the (unique I think) genus 2 curve with an order 5 automorphism. This gives rise to the statement about the points on the line in $\mathbb{P}^2$.

Question 2: Does the statement hold when $C$ contains lines? For example if $C$ is the union of 6 disjoint lines?

The above proof breaks down in this case.