We will use the following relation: For any angle $\theta$,
$$\cos 3\theta = \cos \theta - 4 \cos \theta \sin^2 \theta$$
Let $a, b, c$ be the sides of the triangle and $\alpha, \beta, \gamma$ be the respective angles opposite these sides. Since the triangles $A_{1}BC$, $AB_{1}C$ and $ABC_{1}$ are all congruent to the triangle $ABC$, we have that $\widehat{C_{1}AB_{1}} = 3\alpha$ or $|2\pi - 3\alpha|$, $\widehat{A_{1}BC_{1}} = 3\beta$ or $|2\pi - 3\beta|$, and $\widehat{B_{1}CA_{1}} = 3\gamma$ or $|2\pi - 3\gamma|$.

Applying the Cosine Law to triangle $B_{1}CA_{1}$ yields that
$$A_{1}B_{1}^2 = a^2 + b^2 - 2ab \cos 3\gamma$$
where $R$ is the circumradius of triangle $ABC$, the Cosine and the Sine Laws applied to that triangle yields that
$$2ab \cos \gamma = a^2 + b^2 - c^2$$
and
$$\sin \gamma = \frac{c}{2R}$$
Applying the above to the previous equation yields that
$$\begin{gathered} A_{1}B_{1}^2 = a^2 + b^2 - 2ab \cos 3\gamma = a^2 + b^2 - 2ab \cos \gamma (1 - 4 \sin^2 \gamma) \\ = a^2 + b^2 - (a^2 + b^2 - c^2)(1 - 4 \sin^2 \gamma) \\ = (a^2 + b^2)(1 - 1 + 4 \sin^2 \gamma) + c^2(1 - 4 \sin^2 \gamma) \\ = (a^2 + b^2 - c^2)(4 \sin^2 \gamma) + c^2 = \frac{(a^2 + b^2 - c^2)c^2}{R^2} + c^2 \\ = \frac{c^2}{R^2}(a^2 + b^2 - c^2 + R^2) \end{gathered}$$
Similarly,
$$B_{1}C_{1}^2 = \frac{a^2}{R^2}(b^2 + c^2 - a^2 + R^2)$$
and
$$C_{1}A_{1}^2 = \frac{b^2}{R^2}(c^2 + a^2 - b^2 + R^2)$$
It follows that $A_{1}B_{1} = B_{1}C_{1}$ if and only if
$$c^2(a^2 + b^2 - c^2 + R^2) - a^2(b^2 + c^2 - a^2 + R^2) = 0$$
Factoring the left side yields that
$$(c^2 - a^2)(R^2 + b^2 - a^2 - c^2) = 0$$
The equality of other pairs of sides can be similarly handled. Thus, triangle $A_{1}B_{1}C_{1}$ is equilateral if and only if the following system of three equations is valid:
$$\begin{aligned} & (c^2 - a^2)(R^2 + b^2 - a^2 - c^2) = 0 \\ & (a^2 - b^2)(R^2 + c^2 - b^2 - a^2) = 0 \\ & (b^2 - c^2)(R^2 + a^2 - b^2 - c^2) = 0 \end{aligned}$$
If $a$ is unequal to both $b$ and $c$, then
$$R^2 = a^2 + c^2 - b^2 = b^2 + a^2 - c^2$$
so that $b = c$. Hence, the triangle is isosceles in any case.

Assume that $b = c$. Then from the middle equation, we obtain that
$$(a^2 - b^2)(R^2 - a^2) = 0$$
Therefore, either $a = b$, in which case the triangle is equilateral, or $R = a$ and $\alpha = a / 2R = 1/2$. Therefore, $\alpha = 30^\circ$ or $\alpha = 150^\circ$. Thus, there are three possible sets of angles for the triangle $ABC$: $(60^\circ, 60^\circ, 60^\circ)$, $(30^\circ, 75^\circ, 75^\circ)$ and $(150^\circ, 15^\circ, 15^\circ)$.