Let $A_1$, $B_1$, $C_1$ be the tangency points of the incircle of triangle $ABC$ with the sides $BC$, $CA$, $AB$, respectively.

Since $IA' = IB' = IC'$ it follows that triangles $IA_1A'$, $IB_1B'$, and $IC_1C'$ are congruent. We get $A_1A' = B_1B' = C_1C'$, hence
$$\begin{equation*} \left|(s-b)-\frac{a}{2}\right| = \left|(s-c)-\frac{b}{2}\right| = \left|(s-a)-\frac{c}{2}\right|, \tag{1} \end{equation*}$$
where $a$, $b$, $c$ are the length sides of triangle $ABC$, and $s$ its semiperimeter. The relations (1) are equivalent to
$$\begin{equation*} |c-b| = |a-c| = |b-a|. \tag{2} \end{equation*}$$
From (2), considering all 6 possible orders for $a$, $b$, $c$, it follows $a = b = c$.

The relations $IA' = IB' = IC'$ imply that $I = O_9$, the center of the Euler nine-point circle of triangle $ABC$. Hence $HI = IO$, where $H$ is the orthocenter and $O$ the circumcenter of triangle $ABC$.

But $\widehat{BAH} = \widehat{OAC} = \frac{\pi}{2} - \widehat{B}$, that is $\widehat{HAI} = \widehat{OAI}$. That is triangle $HAO$ is isosceles, hence $AH = AO$.

In similar way, we get $BH = BO$ and $CH = CO$. Since $OA = OB = OC$, it follows $HA = HB = HC$, that is $H = O$, hence triangle $ABC$ is equilateral.

We have
$IA^2 = r^2 + (s-a)^2$, $IB^2 = r^2 + (s-b)^2$, $IC^2 = r^2 + (s-c)^2$,
where $a$, $b$, $c$ are the length sides of triangle $ABC$, $s$ the semiperimeter, and $r$ the inradius.

Applying the Median Theorem in triangle $BIC$, we get
$IA'^2 = \frac{1}{2}(IB^2 + IC^2) - \frac{1}{4}BC^2 = r^2 + \frac{1}{2}[(s-b)^2 + (s-c)^2] - \frac{a^2}{4}$, and similarly
$$\begin{aligned} & IB'^2 = r^2 + \frac{1}{2}[(s-a)^2 + (s-c)^2] - \frac{b^2}{4} \\ & IC'^2 = r^2 + \frac{1}{2}[(s-a)^2 + (s-b)^2] - \frac{c^2}{4} \end{aligned}$$
It follows that $IA' = IB'$ if and only if $IA'^2 = IB'^2$, that is
$$\begin{equation*} (a-b)\left[c - \frac{1}{2}(a+b)\right] = 0 \tag{1} \end{equation*}$$
Also, $IB' = IC'$ if and only if
$$\begin{equation*} (b-c)\left[a - \frac{1}{2}(b+c)\right] = 0 \tag{2} \end{equation*}$$
Relations (1) and (2) hold if and only if $a = b = c$.