How much can we tell about $\det(X)$ if we know $\det(I + X)$? 
*

*What can we tell about $\det(X)$ if we know $\det(I + X)$? Will it
give some kind of bound for $\det(X)$?

*In general, if we know the determinant of matrix $A + X$, where $A$ is a constant matrix, how much can we say about $\det(X)$?


Thank you for the attention.
 A: The determinant is the product of the eigenvalues $$\det({\bf A}) =  \prod_k \lambda_k({\bf A})$$
Then it is well known that all eigenvalues increase by 1 if adding $\bf I$ to any matrix:
$$\det({\bf A+I}) =  \prod_k (\lambda_k({\bf A})+1)$$
Maybe you can rewrite this product into something you are comfortable working with. I do not think this extends in any nice way to adding $\bf X\neq I$.
A: Let $\{\lambda_1,\dots,\lambda_n\}$ be the set of eigenvalues of $X$. Then the set of eigevalues of $I+X$ is $\{1+\lambda_1,\dots,1+\lambda_n\}$. Determinant of any matrix $X$ is equal to product of all eigenvalues of $X$, thus
\begin{align}
\det(X) = \prod_{i=1}^n\lambda_i &&\det(I+X) = \prod_{i=1}^n(1+\lambda_i) \tag{1}
\end{align}
From (1) it is clear that (in general) nothing can be said about $\det(I+X)$.
If all eigenvalues of $X$ are real and positive (i.e. $X$ is positive definite), then $\det(I+X) > \det(X)$.
A: If we are given an invertible matrix $X \in \mathbb R^{n \times n}$ and the following determinant
$$\beta := \det (X + \alpha 1_n 1_n^T)$$
then
$$\beta = \det (X + \alpha 1_n 1_n^T) = \det (X) \cdot \underbrace{\det (I_n + \alpha X^{-1} 1_n 1_n^T)}_{= 1 + \alpha 1_n^T X^{-1} 1_n} = (1 + \alpha 1_n^T X^{-1} 1_n) \cdot \det (X)$$
where we used Weinstein-Aronszajn determinant identity. Thus,
$$\det (X) = \frac{\beta}{1 + \alpha 1_n^T X^{-1} 1_n}$$
A: Nothing To document that "nothing if $d>1$" where $d$ is the dimension is the answer (as said above) :
Claim: For any $\lambda\in \mathbb R$ there exists a matrix $X$ such that $\det X  = \lambda$, while $\det(I+X)=0$.
As a corollary (take $\lambda = \pm n\in\mathbb N$ in the first claim),
Corollary: there exists a sequence $X_n$ such that $\det X_n \to \pm \infty$ while $\det(I+X_n)=0$.
This tells you that nothing general can be said.
Proof. If $d$ is odd take $X$ to be be diagonal matrix with $X_{11}=\lambda$ and $X_{kk}=1$ for $2\leq k\leq d-1$ and $X_{dd}=-1$.  If $d$ is even and greater or equal to $4$, $X_{kk}=1$ for $2\leq k\leq d-2$ and $X_{d-1,d-1}= X_{dd}=-1$. if$d=2$, take $X=\begin{pmatrix} -1 & \lambda \\ -1 & -1\end{pmatrix}$.
Something
On the other hand, it could be that you meant something else than what you asked. For example, what is true is Hadamard's inequality:
$$
\| \det(A+X) -\det(A) \| \leq C(d) \|X\| \max(\|A+X\|^{d-1},\|A\|^{d-1})
$$
