Every inaccessible cardinal is a fixed point of the operation $P$
that assigns to every set $X$ of ordinals the set $P(X)=\{2^{|\alpha|}:\alpha\in X\}\cup\bigcup X$.
On the other hand, every (nonempty) fixed point of $P$ is a strong limit, the least nonempty
fixed point being $\omega$.

Now here is the problem: If you have any operation $R$ on sets of ordinals that is

(1) monotone, i.e., $A\leq B$ implies $R(A)\leq R(B)$, and

(2) continuous, i.e., if $\mathcal A$ is an increasing chain of sets of ordinals, then $R(\bigcup\mathcal A)=\bigcup\{R(A):A\in\mathcal A\}$,

then you can construct fixed points by choosing an increasing sequence $(\alpha_n)_{n\in\omega}$ such that for all $n$, $R(\alpha_n)<\alpha_{n+1}$.
Then $\alpha=\sup_{n\in\omega}\alpha_n$ is a fixed point of $R$, but it is of countable
cofinality and hence, if uncountable, not inaccessible.

If you don't like the fact that I am using functions from sets of ordinals to sets of ordinals instead of functions from ordinals to ordinals, just observe that the "sets of ordinals" formulation is actually more general.
Strong limits are also fixed points of the operation
$\alpha\mapsto\sup\{2^{|\beta|}:\beta<\alpha\}$.

So, if you want inaccesibles as fixed points, you will have to go with operations that are either not monotone or not continuous.
A non-monotone operation that works is this: map every ordinal to the first inaccessible.
(Doesn't charaterize inaccessibles, just the first.)
A non-continuous operation that works is this:

Map every ordinal $\alpha$ to itself if it is inaccessible and to $\alpha+1$ otherwise.
This is of course silly and hence you have to give some more restrictions on the kind of functions that you allow. I think monotone and continuous is natural, but doesn't work.