Unfortunately, this question is wrong.
I will give a counter example, and then show the results I have.
Counter Example
Let $A=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}$, $S=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$, and $\lambda=-1$.
Note that $S$ switches the first basis element with the second, and keeps the third.
From here (or by a simple calculation) it follows that $S^{-1}=S$, and also $SAS^{-1}=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&0\end{pmatrix}$, which means:
$SAS^{-1}=\lambda A$
However, $\lambda^3=(-1)^3=-1$ and $A$ clearly isn't nilpotent.
Alternative Result
Notation:
$A\in M_n(\mathbb{C})$ is a matrix,
$\alpha_1,\ldots,\alpha_n$ are the eigenvalues of $A$ (with multiplicities),
$S\in M_n(\mathbb{C})$ is invertible,
$\lambda\in\mathbb{C}$ is a scalar,
$V_{B,\beta}$ is the eigenspace of $\beta$ with respect to the matrix $B$
and finally $p_B$ is the characteristic polynomial of $B$.
As you noted, in the case $\det A\neq0$ it can be shown that $\lambda^n=1$.
This can be generalized as follows:
Theorem.
Denote by $k$ be the algebraic multiplicity of $0$.
If $SAS^{-1}=\lambda A$ then $\lambda^{n-k}=1$.
Proof.
If $k=n$ the claim is trivial, since $\lambda^0=1$. From here we assume $k<n$
Also, WLOG we assume that the $0$ eigenvalues are last in the list, i.e. $\alpha_1,\ldots,\alpha_{n-k}\neq 0$.
First we prove that $|\lambda|=1$.
We think of multiplying by $\lambda$ geometrically, this operation rotates and stretches the value, but since it takes a finite list to itself, it can't stretch.
Formally, if $|\lambda|>1$, take an eigenvalue with minimal absolute value, its product with $\lambda$ is also an eigenvalue, but it has a bigger absolute value which contradicts the minimality (the case $|\lambda|<1$ is similar).
Thus indeed $|\lambda|=1$, and multiplying by $\lambda$ is equivalent to rotating by some angle.
Now we are moving to prove that $\lambda^{n-k}=1$.
We know that $p_A(x)=\prod_{i=1}^n{(x-\alpha_i)}$, and it is a well known fact that $p_A(x)=p_{SAS^{-1}}(x)$.
Using the lemma below, $p_{\lambda A}(x)=\prod_{i=1}^n{(x-\lambda\alpha_i)}$.
We are given that $SAS^{-1}=\lambda A$.
Altogether $\prod_{i=1}^n{(x-\alpha_i)}=p_A(x)=p_{SAS^{-1}}(x)=p_{\lambda A}(x)=\prod_{i=1}^n{(x-\lambda\alpha_i)}$.
It follows the list $\alpha_1,\ldots\alpha_n$ is equal (up to permutation) to the list $\lambda\alpha_1,\ldots,\lambda\alpha_n$.
Particularly, multiplying an eigenvalue by $\lambda$ gives an eigenvalue.
Geometrically thinking of our eigenvalues will give the desired result.
We should think of them as orbits around the origin, where having multiplicities amounts to two different coinciding orbits.
To put this more formally, for each $1\leq i\leq n-k$, consider $O_i=\{\alpha_i,\lambda\alpha_i,\lambda^2\alpha_i,\lambda^3\alpha_i,\ldots\}$.
Define them with multiplicity, such that two orbits that contain different instances of the same eigenvalue are different.
As we noted, all of the values in $O_i$ are eigenvalues, so $O_i$ must be finite, also it is evident (geometrically and algebraically) that they are of the same length, say $|O_i|=l$. Finally it is evident (with the same picture in mind) that $\lambda^l\alpha_i=\alpha_i$ thus $\lambda^l=1$.
The collection of orbits partitions the list $\alpha_1,\ldots,\alpha_{n-k}$ into $p=\frac{n-k}{l}=$ different orbits, which is a natural number.
Therefore $\lambda^{n-k}=\lambda^{lp}=(\lambda^l)^p=1^p=1$, as required. $\square$
Lemma.
Let $A\in M_n(\mathbb{C})$ and $\lambda\in\mathbb{C}$.
Let $\alpha_1,\ldots,\alpha_n$ be the (not necessarily distinct) eigenvalues of $A$.
Then $\lambda\alpha_1,\ldots,\lambda\alpha_n$ are the eigenvalues of $\lambda A$.
This is equivalent to $p_{\lambda A}=\prod_{i=1}^n{(x-\lambda\alpha_i)}$.
Proof.
Let $v_i\in V_{A,\alpha_i}$, i.e. $Av_i=\alpha_i v_i$.
It follows that $(\lambda A)v_i=\lambda(Av_i)=\lambda(\alpha_i v_i)=(\lambda\alpha_i)v_i$, which means that $\lambda\alpha_i$ is an eigenvalue of $\lambda A$ and $v_i\in V_{\lambda A,\alpha_i}$.
The last paragraph shows that $V_{\lambda A,\lambda\alpha_i}=V_{A,\alpha_i}$ (as the trick works in both directions), and the lemma follows. $\square$
