Your question is not very clearly phrased, which may explain why you didn't get any answers on MSE.
When you say, "spherical curve $\gamma$ whose tangent indicatrix is the same as the original curve, modulo an isometry of $\mathbb{R}^3$", the simple interpretation is this: Assume that $\gamma$ is unit speed, so that $\|\gamma'(t)\|=1$ for all $t$. Then you seem to want to understand the conditions under which one might have
$$
\gamma'(t) = R\,\gamma(t)
$$
for all $t$, where $R:\mathbb{R}^3\to\mathbb{R}^3$ is an isometry of $\mathbb{R}^3$.
Now this can only happen when $\gamma$ parametrizes an arc of a great circle.
For, if $\gamma$ is a unit speed curve on the sphere, then we have the usual result that
$$
\gamma''(t) = -\gamma(t) + \kappa(t)\,\,\gamma(t)\times\gamma'(t),
$$
where $\kappa(t)$ is the (geodesic) curvature of $\gamma$ as a spherical curve.
Thus, $\|\gamma''(t)\|^2 = 1 + \kappa(t)^2$. Meanwhile, if $\gamma'(t) = R\,\gamma(t) = A\,\gamma(t) + b$ where $A$ is a constant orthogonal matrix and $b$ is a constant, then $\gamma''(t) = A\,\gamma'(t)$, which would imply that
$\|\gamma''(t)\|^2 = \|\gamma'(t)\|^2 = 1$, since $A$ is orthogonal. Consequently, we would have to have $\kappa(t)\equiv 0$, which implies that $\gamma$ is indeed a great circle on the unit sphere.
However, you might have wanted to ask whether it was possible for the image of $\gamma$ to be congruent, as a curve in the $2$-sphere, to the image of $\gamma'$. (Again, I'm assuming that $\gamma$ has been parametrized so as to have unit speed.) This is a more interesting question, and somewhat subtle.
Essentially, what this is asking, supposing that $\gamma:I\to S^2$ be a unit speed curve and $I\subset \mathbb{R}$ be an open interval, is whether there exists a function $f:I\to I$ and a rotation $A\in \mathrm{SO}(3)$ such that
$$
\gamma'(t) = A\,\gamma\bigl(f(t)\bigr).
$$
Differentiating this equation, we get
$$
-\gamma(t) + \kappa(t)\,\,\gamma(t)\times\gamma'(t) = \gamma''(t)
= A\,\gamma'\bigl(f(t)\bigr)\,f'(t),
$$
which implies the equation $1+\kappa(t)^2 = f'(t)^2$. Moreover, since $\gamma'$ and $\gamma{\circ}f$ are congruent curves (although not unit speed), they have to have equal geodesic curvature at corresponding points. Using the usual Frenet apparatus to compute this, one sees that this is equivalent to requiring that
$$
\kappa'(t) = f'(t)^3\,\kappa\bigl(f(t)\bigr).
$$
Conversely, if $f:I\to I$ and $\kappa:I\to\mathbb{R}$ are, say, differentiable functions satisfying the conditions
$$
f'(t)^2 = 1+\kappa(t)^2\quad\text{and}\quad
\kappa'(t) = f'(t)^3\,\kappa\bigl(f(t)\bigr),\tag1
$$
then there will exist a unit speed curve $\gamma:I\to S^2\subset\mathbb{R}^3$ with geodesic curvature $\kappa$ such that the curves $\gamma'$ and $\gamma{\circ}f$ are congruent, and such a $\gamma$ will be unique up to isometry.
The functional-differential system (1) is somewhat interesting. It is not obvious that there is a solution $(I, f, \kappa)$ that is nontrivial, i.e., that has $\kappa\not=0$. It's clear that, if $I\subset\mathbb{R}$ is a bounded interval, then $\kappa=0$ is the only possibility. It's also easy to see that if $f$ has a fixed point at $a\in I$, i.e., $f(a) = a$ and $f'(a)= b$ where $|b|>1$, then there is an essentially unique formal power series solution $(f,\kappa)$ in a neighborhood of $a$ with $f(a) = a$ and $f'(a) = b$. (If $(f,\kappa)$ is a solution then $(f,-\kappa)$ is also a solution.) (One can easily reduce to the case $a = 0$ by translating $I$. There cannot be more than one fixed point unless $\kappa\equiv0$ between them.) Unfortunately, it is not clear when this formal power series solution has a positive radius of convergence, though.
Perhaps there is a solution on $I = (0,\infty)$ or $I=\mathbb{R}$ with, say, $f(t)>t$ for all $t$ (i.e., no fixed points), but nothing obvious suggests itself to me.
