Here is the answer to the generalized form of your question in the case of bounded sets. As it turns out, the convexity assumption is unnecessary.
Proposition. Let $A\subseteq\Bbb R^2$ be a bounded set, such that for each pair of distinct points $x,y\in A$ there is a point $z\in A$, such that $x,y,z$ are the vertices of an equilateral triangle. Then $A$ is one of the following:
- empty set,
- a set containing only one point,
- the set of vertices of some equilateral triangle.
Proof. Let $A$ be a set satisfying the hypotheses of the proposition. Suppose $A$ contains more than one point. We will show that then $A$ must be the set of vertices of some equilateral triangle.
First, we shall prove the proposition in the case that $A$ is closed. (In the end we shall show that the general case follows easily from this one.) So, let's assume $A$ is closed, i.e. $A$ contains all its limit points. Then $A$ is compact, so there exist points $a,b\in A$ such that $d(a,b)=\operatorname{diam}A=\sup\lbrace d(x,y)|\text{ }x,y\in A\rbrace,$ where $d$ is the metric in $\Bbb R^2$ and $\operatorname{diam}$ stands for diameter.
Let $c$ be a point such that $a,b,c$ are the vertices of an equilateral triangle. This exists by the hypotheses. Let $r=d(a,b)$ and let $S=\overline{K}(a,r)\cap\overline{K}(b,r)\cap\overline{K}(c,r)$, where $\overline{K}(x,r)$ denotes the closed ball centered at $x$ with radius $r$. Then $A\subseteq S$, because otherwise we would have $\operatorname{diam} A>r$ which is not the case. By the way, $S$ is a Reuleaux triangle:

We already know that $A$ contains the vertices of this Reuleaux triangle. We shall now show that this is all that $A$ contains. We shall first define some more points. Let $a'$ be the reflection of $a$ across the line through $b$ and $c$. Define $b'$ and $c'$ analogously. (As reflections of $b$ and $c$ across the lines opposite to them.) Let $S_{a}:=S\setminus(\overline{K}(b',r)\cup\overline{K}(c',r))\\S_{b}:=S\setminus(\overline{K}(c',r)\cup\overline{K}(a',r))\\S_{c}:=S\setminus(\overline{K}(a',r)\cup\overline{K}(b',r))$ Now, note that every point of $S\setminus\lbrace a,b,c\rbrace$ is contained in at least one of the sets $S_a,S_b,S_c$. So, to complete the proof in the case where $A$ is closed, it suffices to show that $A\cap S_a=\emptyset,A\cap S_b=\emptyset$ and $A\cap S_c=\emptyset$, which we will now do.
Actually, we will only prove the case of $S_a$. The other two can be proved by a completely symmetric argument, so we leave them to the reader. Let $x\in S_a$. Let $u(x)$ be the point that forms an equilateral triangle together with $a$ and $x$, for which the triangle $a,x,u(x)$ is oriented clockwise (i.e. negatively). Let $v(x)$ be the other point that forms an equilateral triangle with $a$ and $x$, i.e. $a,x,v(x)$ is oriented counterclockwise (positively). This defines two functions $u:S_a\to\Bbb R^2$, $v:S_a\to\Bbb R^2$. By definition, $u$ is a rotation around the point $a$ by $-\frac{\pi}{3}$, and $v$ is the rotation around the point $a$ by $\frac{\pi}3$.
By exploiting this geometry, one can now easily see that $u$ rotates the set $S_a$ outside of $S$, i.e. $u(S_a)\subseteq \Bbb R^2\setminus S$, and for $v$ the same holds: $v(S_a)\subseteq \Bbb R^2\setminus S$. But this means that for each $x\in S_a$ the only vertices that could form an equilateral triangle with $x$ and $a$ are outside of $S$, hence not in $A$. But then $x$ cannot be an element of $A$. Here's a picture showing $S_a$ (middle), and its two rotations (the Reuleaux triangle around $S_a$ is $S$, and the two rotations lie outside of $S$):

This completes the proof of the closed case.
In the general case of bounded (not necessarily closed) $A$ with at least two points, we proceed as follows. Let $\overline{A}$ be the closure of $A$. Since $\Bbb R^2$ is such a nice (i.e. first-countable) space, this means that $\overline{A}=\lbrace x\in\Bbb R^2|\text{ there is a sequence } (x_n)_n\text{ with terms in } A\text{, such that } \lim_{n\to\infty}x_n=x\rbrace.$ Now, clearly $A\subseteq\overline{A}$. Furthermore, $\overline{A}$ also satisfies the hypotheses of the proposition. To see this, let $x,y\in\overline{A}$. Then there are sequences $(x_n)_n$ and $(y_n)_n$ in $A$, such that $\lim_{n\to\infty}x_n=x$, $\lim_{n\to\infty}x_n=x$. But since $x_n,y_n\in A$ for each $n$, we can associate a point $z_n\in A$ to each pair $x_n,y_n\in A$ such that $x_n,y_n,z_n$ are the vertices of an equilateral triangle. Because $\overline{A}$ is compact, we can choose a subsequence of $(z_n)_n$ that converges to a point $z\in\overline{A}$. By continuity of the metric, $x,y,z$ again form an equilateral triangle. So, $\overline{A}$ indeed satisfies the hypotheses of the proposition and is therefore the set of vertices of some equilateral triangle. From this it easily follows that $\overline{A}=A$, which concludes the proof. $\square$
This proves among other things, that there is no continuous function $f:[0,1]\to\Bbb R$, whose graph contains for every pair of points a third point that forms an equilateral triangle with them. It proves even more: there is also no such discontinuous function $f:[0,1]\to\Bbb R$. Why? Suppose there is. Then it must be bounded, since, otherwise its graph would contain two points $x,y$ such that $d(x,y)>100$. There would then have to be a third point on the graph whose first coordinate would lie outside of $[0,1]$, which is not the case. But for bounded sets, the proposition applies. The same applies for any function $f:B\to\Bbb R$, where $B\subseteq\Bbb R$ is a bounded set.
The situation might be more interesting with functions $f:\Bbb R\to\Bbb R$, however. But in this case again, there is no such bounded function. (Same argument as in the previous paragraph.) It might be possible to construct something unbounded and probably horribly discontinuous, though, but I haven't thought much about that.