In ECC, the private key $k$ is a scalar and a public key is $[k]G$, this is scalar multiplication defined as
\begin{align}
[k]:& E \to E\\
&P\mapsto [k]G=\underbrace{G+G+\cdots+G}_{\text{$k$ times}}.\end{align}
As we can see, it is a short definition of multiple additions of the same element. Since in ECC we work in finite fields $\mathbf{F}$, the elements of the curve are finite and actually forms an abelian group with an identity $\mathcal{O}$, this can be the point of infinity or not, depending on the curve definition. If the order of the point $G$ is $t$ ( the $t$ can be equal to the order of the EC group or a divisor of it, i.e. $ord(G)\mid \#|E(\mathbf{F})|$) then
$$\underbrace{G+G+\cdots+G}_{\text{$t$ times}} = [t]G= \mathcal{O}$$
Inverse of private key
Now if you want to the inverse of $k^{-1}$ it is just $t-k$. You can see this from this
$$[k]G + [t-k]G = [k+t-k]G = [t]G = \mathcal{O}$$
Finding the base point only with private and public key
The question has little meaning since the base point comes with a standard and it is publicly known for all parties. To able to calculate any arithmetic, you already must know the Elliptic curve construction parameters.
let's consider you all know but the basepoint $G$ and formalizing ;
find $x$ such that $[x]P = G$ where $P = [k]G$ given $P$ and $k$.
We will use this identity on the scalars;
$$[k]([a]P) = [ak]P$$
$$[k]([a]P)=\underbrace{ \underbrace{P+P+\cdots+P}_{\text{$a$ times}} + \cdots + \underbrace{P+P+\cdots+P}_{\text{$a$ times}} }_{ \text{$k$ times}}$$
Since the scalars are working a modulus, i.e.
$$[k]P = [k \bmod t]P$$ where $t$ was the order of the $G$. So, as given by the corspini's answer if the multiplicative inverse of $k$ over the group of order $t$ then we can use it to find $G$. In ECC, we use usually use prime order subgroups so inverses exist in them.
Let $k'\cdot k \equiv 1 \bmod t$ and thake $P = [k]G$ now add $k'$ times
$$[k' \bmod t]P = [k'([k]]G) = [k' \cdot k \bmod t]G = G$$