Let $M_n = M_n(\Bbb{C})$ be the algebra of $n\times n$ complex matrices and let $\|\cdot\|$ be the operator norm. Consider the map $$f : M_n \to M_n, \qquad f(X) = X + \|X-I\|I.$$ I'm interested in the image of this map.
First I noticed that the image seems to be contained in the group of invertible matrices $GL(n)$. Namely, for all $X \in M_n$ we need to show that $X + \|X-I\|I$ is invertible. By plugging in $X+I$ it is equivalent to show that $X + (1+\|X\|)I$ is invertible for all $X \in M_n$. But this is true since the spectrum $\sigma(X)$ of $X$ is contained in the closed ball of radius $\|X\|$ around the origin so clearly $-(1+\|X\|) \notin \sigma(X)$.
For $Y \in GL(n)$ we can try to find $X \in M_n$ such that $f(X) = Y$. Clearly it has to be of the form $X = Y - \alpha I$ for some scalar $\alpha \ge 0$. Plugging in we have $$Y-\alpha I + \|(Y-\alpha I) - I\| = f(Y-\alpha I) = Y \implies \alpha = \|Y - (\alpha + 1)I\|.$$
The question can be reformulated to: for which $Y \in GL(n)$ exists $\alpha \ge 0$ such that $\alpha = \|Y - (\alpha + 1)I\|$?
It is easy to see that there is no such $\alpha$ for $Y = \beta I$ when $\beta \in \langle -\infty, 1\rangle$. So maybe the image is $$GL(n)\setminus \{\beta I : \beta \in \langle -\infty, 1\rangle\}?$$
Any partial information about the image is welcome.