Induced morphism on automorphism groups in terms of the original map. I recently came across the following problem and I haven't solved it yet.
Let $f:X\to Y$ be a continuous map and let $p:\tilde{X}\to X$ and $q:\tilde{Y}\to Y$ be the universal covering spaces. There are canonical isomorphisms $\pi_1(X)\simeq Aut(\tilde{X},p)$ and $\pi_1(Y)\simeq Aut(\tilde{Y},q)$, and there is an induced map $f_*:\pi_1(X)\to \pi_1(Y)$. If we conjuguate $f_*$ by the canonical isomorphisms, we obtain a map 
$$F: Aut(\tilde{X},p)\longrightarrow Aut(\tilde{Y},q).$$

Question: What is $F$ in terms of $f$ ?

My guess: We denote by $\tilde{f}:\tilde{X}\to \tilde{Y}$ a lift of $f$. If $g\in Aut(\tilde{X},p)$, then $F(g)$ is the unique element of $Aut(\tilde{Y},q)$ satisfying $$\tilde{f}\circ g=F(g)\circ \tilde{f},$$
i.e we have the following commutative diagram
$$\require{AMScd} \begin{CD} \tilde{X} @>{g}>> \tilde{X}\\ @V{\tilde{f}}VV @VV{\tilde{f}}V\\ \tilde{Y} @>>F(g)> \tilde{Y} \end{CD}$$
Of course the lift of $f$ and the canonical isomorphims depend of a base-point but I don't write it explicitly to make my question easier to read.
Am I right? I am not really looking for a proof right now I just want to know if my guess is right.
Thanks in advance !
Edit: after some thinking I think it is better to be precise about the base point. So we fix $x_0\in X$, $y_0=f(x_0)\in Y$, $\tilde{x_0}$ in the fiber $p^{-1}(\{x_0\})$ and $\tilde{y_0}$ in the fiber $q^{-1}(\{y_0\})$. We denote $\tilde{f}$ the unique lift of $f\circ p$ by $q$ such that $\tilde{f}(\tilde{x_0})=\tilde{y_0}$, which gives the commutative diagram 
$$\require{AMScd} \begin{CD} (\tilde{X},\tilde{x_0}) @>{\tilde{f}}>> (\tilde{Y},\tilde{y_0})\\ @V{p}VV @VV{q}V\\ (X,x_0) @>f>> (Y,y_0) \end{CD}$$
Let $\varphi :\pi_1(X,x_0)\xrightarrow{\sim} Aut(\tilde{X},p)$ be the isomorphism sending $[\gamma]$ to the unique covering transformation $g$ such that 
$$g(\tilde{x_0})=\tilde{x_0}\cdot[\gamma].$$
Similarly let $\psi :\pi_1(Y,y_0)\xrightarrow{\sim} Aut(\tilde{Y},q)$ be the isomorphism sending $[\gamma]$ to the unique covering transformation $g^{\prime}$ such that $$g^{\prime}(\tilde{y_0})=\tilde{y_0}\cdot[\gamma].$$
I am interested in $\psi\circ f_*\circ \varphi^{-1}$. 
 A: With the help of a friend I found out that my guess was correct. 
Let $g\in Aut(\tilde{X},p)$, we want to show 
$$\begin{equation}\tag{1}\tilde{f}\circ g=F(g)\circ \tilde{f}\end{equation}$$
We have that $$q\circ (\tilde{f}\circ g )= f\circ p\circ g=f\circ p$$ and 
$$q\circ (F(g)\circ \tilde{f})=q\circ \tilde{f}=f\circ p.$$
If we assume that $X$ is connected (which is the case because if not it wouldn't make sens to talk about fundamental group), then from uniqueness of the lift of an application, $(1)$ is true if and only if it is true at one point. 
We now prove that $(1)$ is true at $\tilde{x_0}$. Let $[\gamma]\in \pi_1(X,x_0)$ such that $$g(\tilde{x_0})=\tilde{x_0}\cdot[\gamma]$$ 
and take $\tilde{\gamma}$ a lift of $\gamma$ starting at $\tilde{x_0}$ (which by definition implies $\tilde{\gamma}(1)=\tilde{x_0}\cdot[\gamma]=g(\tilde{x_0})$.
By definition of $F$ we have $$F(g)(\tilde{y_0})=\tilde{y_0}\cdot[f\circ\gamma].$$
Then $$\begin{array}{rcl}
F(g)\circ \tilde{f}(\tilde{x_0})&=&\tilde{f}(\tilde{x_0})\cdot[f\circ\gamma]\\
&=&\text{endpoint of a path lifting }f\circ\gamma\text{ starting at }\tilde{f}(\tilde{x_0})\\
&=& \tilde{f}\circ\tilde{\gamma}(1)\\
&=& \tilde{f}\circ g(\tilde{x_0}).
\end{array}$$
