Cartesian dissimilarity of a function $\ f:A^3\to A^3\ $ and its inverse Let $\ A\ $ be an arbitrary set. Let $\ |A|>1\ $ (to avoid triviality). Let each of the functions $\ f_k:A^{\{1\ 2\ 3\}}\to A\ $ depend on all three arguments for $\ k=1\ 2\ 3,\ $ while each of the functions $\ g_k:A^{\{1\ 2\ 3\}}\to A\ $ does not depend on $k$-th variable, for each $\ k=1\ 2\ 3.$
I'll explain "dependent/independent" at the end of this note. It'll be preceded by "diagonal product $\ \triangle\ $ of functions (here, of three of them).
Assume also that
$$ f:=f_1\triangle f_2\triangle f_3\ \ \text{and}
   \ \ g:=f_1\triangle f_2\triangle f_3\, :\,A^3\to A^3 $$
are inverse one to another. There are (historic) examples when $\ A\ $ is countable or of cardinality continuum, and for all infinite cardinals except, possibly, for the weird ones. There are examples (by analogy) whenever $\ |A|\ $ is finite and odd; there should be some examples when $|A|$ is not a power of $2$, but -- I conjecture -- never when it is:
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CONJECTURE:  cardinality $\ |A|, $ when it's finite, is not a (non-trivial) power of $\ 2\ $ (is different from $\ 2^k\ $ for any natural $k=1\ 2\ \ldots$).
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EXAMPLES:
Let $\ A = \Bbb Q\ $ or $\ \Bbb R\ $ or $\ \Bbb Z_n\ $ for arbitrary odd $\ n>1.\ $ Define:


*

*$\ f_1(a\ b\ c)\ :=\ b+c-a $

*$\ f_2(a\ b\ c)\ :=\ a+c-b $

*$\ f_2(a\ b\ c)\ :=\ a+b-c $
and


*

*$\ g_1(a\ b\ c)\ :=\ \frac{b+c}2 $

*$\ g_2(a\ b\ c)\ :=\ \frac{a+c}2 $

*$\ g_2(a\ b\ c)\ :=\ \frac{a+b}2 $
Then, let $\ f\ $ and $\ g\ $ be defined as above. This establishes the odd cardinality case.
Remark:  Every finite abelian group $\ X\ $ of odd order
($\ |X|\, $ -- odd) will do in place of $\ Z_n\,$ (with odd $n$).
The finite even cardinalities different from $\ 2^n\ (n\in\Bbb N),\ $ present a mixed story in an analogy to other combinatorial themes which admit exotic examples. This would be the complementary conjecture -- the mixed picture.
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Diagonal product of functions (or morphisms)
Consider set $\ X\ $ and sets $\ Y_q\ $ and functions
$\ f_q:X\to Y_q\ (q\in Q).\ $ Then, the diagonal product
$\ f:=\triangle_{q\in Q} f_q :X\to\prod_{q\in Q} Y_q\ $
is given by:
$$ \forall_{q\in Q}\quad \pi_q\circ f\ := f_q $$
i.e.
$$ \forall_{q\in Q}\,\forall_{x\in X}\quad (f(x))(q)\ := f_q(x) $$
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Dependent / independent
Let $X\ Y\ T\ $ be arbitrary sets, and $\ s\in T.\ $ Elements $\ x\in X^T\ $
are functions $\ x: T\to X$.
A function $\ f:X^T\to Y\ $ does not depend on (is independent of) variable
$\ s\ \Leftarrow:\Rightarrow$
$$ \exists_{f_s\in X^{T\setminus\{s\}}}\,\forall_{x\in X^T}\quad
     f(x)=f_s(x|T\setminus\{s\}) $$
Otherwise, $\ f\ $ depends on variable $\ s,\ $ i.e.
$$ \exists_{w\ x\,\in\,X^T}\ \ (\, w|T\!\setminus\!\{s\}\,=\,
      x|T\!\setminus\!\{s\}\quad\text{and}\quad f(w)\ne f(x) \,) $$
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PS. In the style of Q&A, I've provided only the special case of the general question about the number of independent variables of a function and its inverse, $\ f\ $ and $\ g.\ $ Actually, we want to know the whole structure of sets of independent variables for $\ f\ $ and $\ g.\ $ This question is as fundamental as it goes, hence it belongs to the theory of the Foundations of Mathematics. (Please, someone reattach the related tag to my question).
 A: $\newcommand{\F}{\mathbb{F}}$
Your conjecture is false, I will construct a counterexample for all powers of two $2^n$ with $n \ge 2$. 
Let us identify $A$ with $\F_{2^n}$. An example from OP corresponding to the odd numbers, is a linear mapping. Our functions also would be linear. Let $G$ be a matrix constructed from the functions $g_1, g_2, g_3$, as rows and $F$ be its inverse, constructed of the functions $f_1, f_2, f_3$. Then we want the following things: matrix $G$ has zeroes on the diagonal and matrix $F$ does not have zero entries. If we start with the matrix $G$ with the zeroes on the diagonal then the elements of $F$ are product of some two elements of $G$, divided by the determinant. So as long as all non-diagonal elements of $G$ are non-zero and the determinant is non-zero we won. This can be achieved for all $n \ge 2$ by, for example, the following construction:
Let $a \in \F_{2^n}$, $a\ne 0, 1$ and consider functions 
$$g_3(x, y, z) = x + y,$$
$$g_2(x, y, z) = x + z,$$
$$g_1(x, y, z) = y + az.$$
then the functions $f$ are
$$f_1(x, y, z) = \frac{1}{a+1}(ay-x+z),$$
$$f_2(x, y, z) = \frac{1}{a+1}(-ay+az+x),$$
$$f_3(x, y, z) = \frac{1}{a+1}(y+x-z).$$
Since $a\ne 1$ we have $a+1\ne 0$ (we are in the field of characteristic two) and so it is a working example.
For the case $N = 2^kM$, $M$ odd, $M > 1$ we can just copy $2^k$ times the construction for $M$ from the OP. It leaves only the cases $N = 1, 2$. For $N = 1$ there are obviously no solutions and for $N = 2$ one can simply bruteforce all the potential cases (as was done by Ycor in the comments while I was writing my answer).
EDIT here is a simpler construction which doesn't need finite fields but only modular arithmetics and work for all $m \ge 3$. We will work in $\mathbb{Z}/m\mathbb{Z}$. Consider $$g_3(x, y, z) = x +y,$$ $$g_2(x, y, z) = x + z,$$ $$g_1(x, y, z) = y-2z.$$ Then the functions $f$ are
$$f_1(x, y, z) = -z+2y+x,$$
$$f_2(x, y, z) = 2z-2y-x,$$
$$f_3(x, y, z) = z-y-x.$$
