This is a coda to the given answer. Your problem is a special case of the classical one of reconstructing a triangle from parts, in this case from special points. As suggested by the above, one requires three of these, the simplest example being that of the vertices.
You are considerng the case where the points are triangle centres but there are many other interesting ones. To my knowledge, such problems were first considered by Euler and were revived in the last century by Wernick (“A triangle construction from three located points”, Math. Mag. 55 (1982) 227-230). They have received some attention recently in the context of automatic proofs in elementary geometry. There is a systematic approach to such problems and we sketch it briefly. Suppose we are given as data the three special points (in your case three triangle centres) $A_1$, $B_1$, and $C_1$. We now consider an arbitrary triangle $ABC$ and determine when its special points are the above three. Wlog we can assume that the vertices are $(0,0)$, $(c,0)$ and $(p,q)$. We calculate the three special points of this triangle—call them $A_2$, $B_2$ and $C_2$ (they depend on $p$, $q$ and $c$). We then consider the three equations
$|B_2C_2|^2=|B_1C_1|^2$, etc, in the variables $p$, $q$ and $c$. Of course, many things can happen, depending on the specific situation, but in general you will have an essentially unique solution and this solves your problem. For simple cases this can be done (and is fun to do) by hand, but in general it is rather tedious or even impossible. It is, however, easy to write a little programme, say with Mathematica, to mechanise the computations. We illustrate this with the case where the data triangle has three triangle centres as vertices. More precisely we suppose that they are determined by three triangle centre functions $f$, $g$ and $h$ of the side lengths $a$, $b$ and $c$ of $ABC$ (we are using the terminology of the Encyclopedia of Triangle Centers—easy to find online). In order to simplify the notation, we assume that these functions are homogeneous and satisfy $f(a,b,c)+f(b,c,a)+f(c,a,b)=1$.
Then the required equations are
$$[c(g(b,c,a)-h(b,c,a))+p(g(c,a,b)-h(c,a,b))]^2+q^2(g(c.a.b)-h(c,a,b))^2=|B_1C_1|^2$$ and its natural cyclic permutations. (These involve five variables—the side lengths and $p$ and $q$ from the coordinates of the vertices of $ABC$ but they can be reduced to equations in three variables, $(a,b,c)$ or $(p,q,c)$, by using the relationships $p^2+q^2=b^2, (p-c)^2+q^2=a^2$). The case considered by Euler is that of the incentre, circumcentre and orthocentre and the corresponding $f$, $g$ and $h$ can be found in the encyclopedia referred to above. Since there are going on 50,000 triangle centres documented we are talking about a potential of more than $40,000^3$ results.
The seeding article of Euler is in vol. 26 of his Opera Omnia, pp. 139-157 (“A simple solution of some very difficult geometrical problems”—my translation from the latin).