Consider a function $f \colon U \to X$, if $f$ is the identity when acting on $X$, then $df_x$ is also the identity for all $x \in X$? While doing an independent project, I ran into this question. Let $X$ be a $k$-dimensional compact manifold in $\mathbb{R}^n$ and let $U \subset \mathbb{R}^n$ be an open set such that $X \subset U$. As in the title, consider a smooth function $f \colon U \to X$ such that $f(x) = x$ for all $x \in X$. Then, is it true that $df_x$ is the identity matrix?

My first guess is that it is not always true. For this, recall what a lefschetz map is. The latter is a function such that on its fixed points $x$ (i.e. $x \in U$ such that $f(x) = x$) $df_x$ does not have $+1$ as an eiganvalue, in other words we do not have $df_x(v) = v$ for all $v$ in $T_x(X)$. So, if the above were true then we would not have lefschetz maps. Is this reasoning correct? Or this depends on the nature of $f$ or $X$?

Any help or advice would be greatly appreciated.
Let $v\in T_{x}U$ and take a curve $c: I \longrightarrow U$ such that $c(0) = x$ and $\dot{c}(0) = v \in T_{x} U \subset T_{x}X$. Like $c(t) \in U$ then $(f \circ c)(t) = c(t)$ and from here
\begin{equation}
df_{x}(v) = df_{x}(\dot{c}(0)) = \left. \frac{d}{dt}(\ f \circ c\ )(t) \right|_{t=0} =  \left. \frac{d}{dt}(\ c(t)\ ) \right|_{t=0} = \dot{c}(0) = v
\end{equation}
Therefore $df_{x}$ is the identity linear transformation.
Let $v\in T_{x}U$ and take a curve $c: I \longrightarrow U$ such that $c(0) = x$ and $\dot{c}(0) = v \in T_{x} U \subset T_{x}X$. Like $c(t) \in U$ then $(f \circ c)(t) = c(t)$ and from here
\begin{equation}
df_{x}(v) = df_{x}(\dot{c}(0)) = \left. \frac{d}{dt}(\ f \circ c\ )(t) \right|_{t=0} =  \left. \frac{d}{dt}(\ c(t)\ ) \right|_{t=0} = \dot{c}(0) = v
\end{equation}
Therefore $df_{x}$ is the identity linear transformation.