As noted in the comment, the notation is common for right actions of a group, and the group of automorphisms acts on the group.
Recall that given a group $G$ and a set $X$, a right action of $G$ on $X$ is a function $X\times G\to X$, denoted with infix notation as $x\cdot g$, such that (i) For all $x\in X$, $x\cdot e = x$; and (ii) For all $x\in X$, $g,h\in G$, $x\cdot(gh) = (x\dot g)\cdot h$.
One can use exponential notation: $x^g = x\cdot g$. Then you have $x^e = x$, and $(x^g)^h = x^{gh}$.
This notation is also common when the action of $G$ on itself (or on a normal subgroup) by conjugation, defining $x^g = g^{-1}xg$.
Now let $K$ be a group, and let $G$ be the group of all automorphisms of $K$. We can let $K$ act on $G$ by letting $x\cdot f = f(x)$ for all $x\in X$, $f\in K$. Writing composition in $K$ from right to left, so that $fg$ means "$f$ first, $g$ second", we have that this is a right action.
(Using functions on the right is common in ring theory, since then a module homomorphism would satisfy a homogeneity rule that reads a "$(am)f = a(mf)$"; for a while during the 60s, some group theorists adopted this notational convention by putting morphisms on the right; e.g., Hanna Neumann's Varieties of Groups has functions on the right. It also has the advantage that $f\circ g$ means "$f$ composed with $g$", i.e., $f$ first, $g$ second, instead of meaning the usual $g$ composed with $f$.)
Now, in fact, you can even realize this action of $\mathrm{Aut}(G)$ on $G$ as conjugation: the holomorph of a group $G$ is the semidirect product $G\rtimes \mathrm{Aut}(G)$, where the action is by $x\cdot f = x^f$.
For reference the number $1707 \uparrow \uparrow 1783$ uses Knuth's up-arrow notation for tetration. For every $n\in\Bbb{N}$, the number $1707\uparrow\uparrow n$ is the $n$-th term in the sequence $(s_n)_{n\in\Bbb{N}}$ defined by $s_0=1$ and $s_{n+1}=1707^{s_n}$.
To determine the remainder of $s_{1783}=1707 \uparrow \uparrow 1783$ after division by $10^{17}$ we repeatedly use Euler's theorem, which tells us that if $s_{1782}\equiv t_0\pmod{\varphi(10^{17})}$ then
$$s_{1783}=1707^{s_{1782}}\equiv1707^{t_0}\pmod{10^{17}}.$$
Repeating the same argument we see that, if $s_{1782-k}\equiv t_k\pmod{\varphi^k(10^{17})}$ then
$$s_{1783-k}=1707^{s_{1782-k}}\equiv1707^{t_k}\pmod{\varphi^{k-1}(10^{17})}.$$
Of course for positive integers $a$ and $b$ we have $\varphi(2^a5^b)=2^{a+1}5^{b-1}$,
so
$$\varphi^k(10^{17})=\begin{cases}
2^{17+k}5^{17-k}&\text{ if }k\leq17,\\
2^{51-k}&\text{ if }17\leq k\leq51,\\
1&\text{ if }k\geq51\end{cases},$$
and hence we may take $t_{50}=1$ as $\varphi^{50}(10^{17})=2$ and $s_{1732}\equiv1\pmod{2}$. As $\varphi^{49}(10^{17})=4$ it follows that
$$s_{1733}=1707^{s_{1732}}\equiv1707^1\equiv3\pmod{4},$$
and so we may take $t_{49}=3$. Repeating then yields
\begin{eqnarray*}
t_{49}=s_{1734}&=&1707^{s_{1733}}&\equiv&1707^3&\equiv&3^3\equiv3&\pmod{8},\\
t_{48}=s_{1735}&=&1707^{s_{1734}}&\equiv&1707^3&\equiv&11^3\equiv3&\pmod{16},\\
\vdots\\
t_{18}=s_{1765}&=&1707^{s_{1765}}&\equiv&1707^{5418148179}&\equiv&14008082771&\pmod{2^{34}},
\end{eqnarray*}
and now repeat a few more times to compute $t_0=s_{1783}\pmod{10^{17}}$.
Best Answer
(Replacing my earlier comments to make a proper answer)
Suggestion 1: In the tetration forum we have partly used $\exp^{\circ h}_b(x)$ (the little circle indicating function-composition instead of powers or instead of derivation) for the general iteration $x \to b^x$ to the iteration-(h)eight $h$ of the exponentialtower. I myself use sometimes $\text{T}^{\circ h}_b(x)$ for shortness and $\text{U}^{\circ h}_b(x)$ for the decremented exponentiation $x \to b^x−1 $.
Suggestion 2: In many articles I've also seen the simple solution to use the index-notation. So $z_0$ for the initial value , $z_1=b^{z_0}$ then $z_h=b^{z_{h−1}}$ for the $h$'th iteration (exponentialtower of (h)eight $h$) and $z_\infty$ if that limit exists. In articles the base $b$ is mostly a fixed parameter over a lot of formulae and algebraic derivations so I'd prefer such a notation which allows to omit this reference to $b$ to reduce redundancy in notation. (I find this unbeatable concise - unfortunately the indexing-notation indicates many things in math so I use this only when I'm well sure it is not obfuscating my line of discussion/derivation/definition)